ELEMENTARY   CALCULUS 


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THE  MACMILLAN  COMPANY 

NEW  YORK    •    BOSTON   •    CHICAGO  •    DALLAS 
ATLANTA   •    SAN    FRANCISCO 

MACMILLAN  &  CO.,  Limited 

LONDON   •    BOMBAY   •    CALCUTTA 
MELBOURNE 

THE  MACMILLAN  CO.  OF  CANADA,  Ltd. 

TORONTO 


.ELEMENTARY  CALCULUS 


BY  »• 

WILLIAM    F.   OSGOOD,   Ph.D.,   LL.D 

PERKINS    PROFESSOR    OF    MATHEMATICS 
IN    HARVARD    DNIVERSITY 


Welti  gotfc 

THE   MACMILLAN    COMPANY 

1921 

All  rights  reserved 


Copyright,  192], 
By  THE  MACMIU.AN  COMPANY 


Set  jp  ancj  electrotyped.     Published  January,  1921. 


Xortoooa  litrss 

J.  S.  Cushing  Co.  —Berwick  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


PREFACE 

The  object  of  this  book  is  to  present  the  elements  of  the 
Differential  Calculus  in  a  form  easily  accessible  for  the  under- 
graduate. It  is  possible,  from  the  very  beginning,  to  illustrate 
the  ideas  and  methods  of  the  Calculus  by  means  of  applications 
to  physics  and  geometry,  which  the  student  can  readily  grasp, 
and  which  will  seem  to  him  of  interest  and  value.  To  do  this, 
the  stress  in  the  illustrative  examples  worked  in  the  text  must 
be  laid  first  of  all  on  the  thought  which  underlies  the  method 
of  solution,  in  distinction  from  the  exposition  of  a  process,  re- 
duced in  the  worst  teaching  to  rules,  whereby  the  answer  can 
be  obtained.  The  treatment  of  maxima  and  minima,  Chapter 
III,  §§  2,  3,  and  curve  tracing,  Chapter  III,  §  5  and  Chapter  VII, 
§  10,  will  serve  to  show  what  is  here  meant. 

It  is,  however,  also  essential  that  the  student  receive  thorough 
training  in  the  formal  processes  and  the  technique  of  the  Cal- 
culus, and  this  side  has  been  treated  with  care  and  complete- 
ness. Note,  for  example,  the  differentiation  of  composite 
functions  in  Chapter  II,  §  8,  and  the  exposition  of  the  use  of 
differentials  in  differentiating  in  Chapter  IV,  §§  4,  5. 

An  important  application  of  the  graphical  methods,  with 
which  the  Calculus  is  so  intimately  associated,  is  that  of 
solving  approximately  numerical  equations  which  do  not 
come  under  the  standard  rules  of  algebra  and  trigonometry. 
Hitherto,  however,  little  attempt  has  been  made  to  present 
this  subject,  simple  as  it  is,  in  any  systematic  and  elementary 
manner.  In  Chapter  VII  the  common  methods  in  use  by 
physicists  and  others  who  apply  the  Calculus  are  set  forth 
and  illustrated  by  simple  examples. 

The  book  might  have  included  a  brief  treatment  of  curva- 
ture   and    evolutes,   and    the    cycloid.      But    probably    most 

v 

0991 


VI  PREFACE 

teachers  of  the  Calculus  will  prefer  to  take  up  integration 
next,  and  so  the  closing  chapter  is  devoted  to  the  last  of  the 
elementary  functions,  the  inverse  trigonometric  functions,  with 
special  reference  to  their  one  great  application  in  the  elements 
of  mathematics,  namely,  their  application  to  integration. 

The  book  is  so  written  that  it  can  be  adapted,  if  desired,  to 
an  abridged  course,  in  which,  after  the  fundamentals  of  the 
first  three  chapters  have  been  covered,  any  of  the  remaining 
topics  can  be  treated  briefly,  and  thus  a  wide  scope  in  subject 
matter  is  possible,  even  when  the  time  is  short. 

Cambridge,  Massachusetts, 
January,  1921. 


CONTENTS 


CHAPTER   I 
INTRODUCTION 

1.  Functions      ....... 

2.  Continuation.     General  Definition  of  a  Function 


1 
10 


CHAPTER   II 

DIFFERENTIATION    OF   ALGEBRAIC    FUNCTIONS. 
THEOREMS 

1.  Definition  of  the  Derivative  . 

2.  Differentiation  of  xn 

3.  Differentiation  of  a  Constant 

4.  Differentiation  of  ^x     . 

5.  Three  Theorems  about  Limits.     Infinity 

6.  General  Formulas  of  Differentiation 

7.  General  Formulas  of  Differentiation,  Continued 

8.  General  Formulas  of  Differentiation,  Concluded 

9.  Differentiation  of  Implicit  Algebraic  Functions 


GENERAL 


13 

16 
20 
21 
22 
29 
32 
35 
39 


CHAPTER   III 
APPLICATIONS 

1.  Tangents  and  Normals 

2.  Maxima  and  Minima    . 

3.  Continuation  :  Auxiliary  Variables 

4.  Increasing  and  Decreasing  Functions 

5.  Curve  Tracing       .... 

6.  Relative  Maxima  and  Minima.     Points  of  Inflection 

7.  Necessary  and  Sufficient  Conditions      .         .         . 

8.  Velocity ;  Rates 


46 
49 
53 
60 
64 
67 
71 
72 


Vlll 


CONTENTS 


CHAPTER    IV 

INFINITESIMALS    AND    DIFFERENTIALS 

1.  Infinitesimals 

2.  Continuation.     Fundamental  Theorem 

3.  Differentials  ...... 

4.  Technique  of  Differentiation 

5.  Continuation.     Differentiation  of  Composite  Functions 


CHAPTER   V 
TRIGONOMETRIC   FUNCTIONS 

■'     1.  Radian  Measure    ..... 

2.  Differentiation  of  sin  x  . 

3.  Certain  Limits 

4.  Critique  of  the  Foregoing  Differentiation 

5.  Differentiation  of  cost,  tan  x,  etc. 

6.  Shop  Work    ...... 

7.  Maxima  and  Minima    .... 

8.  Tangents  in  Polar  Coordinates 

9.  Differential  of  Arc 
10.  Rates  and  Velocities      .... 


CHAPTER    VI 

LOGARITHMS   AND    EXPONENTIALS 


1.  Logarithms    .... 

2.  Differentiation  of  Logarithms 

i 

3.  The  Limit  lim  (l+t)'     . 

4.  The  Compound  Interest  Law 

5.  Differentiation  of  ex 

6.  Graph  of  the  Function  x" 

7.  The  Formulas  of  Differentiation  to  Date 


CHAPTER    VII 
APPLICATIONS 

1.  The  Problem  of  Numerical  Computation 

2.  Solution  of  Equations.     Known  Graphs 


CONTENTS 


IX 


3.  Interpolation         ....  .         . 

4.  Newton's  Method  ...  .         . 

5.  Direct  Use  of  the  Tables       ...... 

6.  Successive  Approximations    ...... 

7.  Arrangement  of  the  Numerical  Work  in  Tabular  Form 

8.  Algebraic  Equations      ....... 

9.  Continuation.     Cubics  and  Biquadratics 

10.  Curve  Plotting      ...... 

CHAPTER   VIII 
THE    INVERSE    TRIGONOMETRIC    FUNCTIONS 


PAGE 

170 
172 
176 
180 
184 
187 
189 
19 


5 


1. 

Inverse  Functions          ...... 

.     206 

2. 

The  Inverse  Trigonometric  Functions  . 

.     209 

3. 

Shop  Work    ........ 

.     215 

4. 

Continuation.     Numerical  Computation 

.     218 

5. 

Applications.         ....... 

.     221 

'. 


CALCULUS 

.     CHAPTER   I 
INTRODUCTION 

The  Calculus  was  invented  in  the  seventeenth  century  by 
the  mathematician,  astronomer,  and  physicist,  Sir  Isaac  Newton 
in  England,  and  the  philosopher  Leibniz  in  Germany.  The 
reaction  of  the  invention  on  geometry  and  mathematical  physics 
was  most  important.  In  fact,  by  far  the  greatest  part  of  the 
mathematics  and  the  physics  of  the  present  day  owes  its 
existence  to  this  invention. 

1.  Functions.  The  word  function,  in  mathematics,  was 
first  applied  to  an  expression  involving  one  or  more  letters 
whii-h  represent  variable  quantities ;  as,  for  example,  the 
expressions 

(a)  Xs,        2a?  —  3x  +  l; 

(b)  Vx,         Vaj  —  x- : 


(<0 


xy  ax  +  by 


a  +  x  x2  +  y2'         Vx2  +  y2  +  z2' 

(d)  sin  x,         log  x,         tan-1  x. 

In  the  second  example  under  (/>),  two  letters  enter;  but  a 

is  thought  of  as  chosen  in  advance  and  then  held  fast,  x  alone 

being  variable.     A  quantity  of  this  kind  is  called  a  constant. 

Thus 

ax  +  b 

is  a  function  of  x  which  depends  on  two  constants,  a  and  b. 

l 


2  CALCULUS 

Such  expressions  are  written  in  symbolic,  or  abbreviated, 
form  as  f(x),  f(x,  y)  (read  :  "/  of  x,"  "foi  x  and  y"  etc.); 
other  letters  in  common  use  being  F,  <f>,  <i>,  etc.*  Thus  the 
equation 

(1)  /(,■)  =  2,^ -:u-  +  i 

defines  the  function /(a;)  in  tin-  presenl  case  t«>  be  2a?  -.'5x  +  l. 

Again, 

(2)  <f>{.r.  y,  -. .  =  *•  +  if  +  z- 

is  an  equation  defining  tin-  function  </>  .<•.  y,  z)  as  .»•-  +  >/-  -f  z2. 

We  shall  be  concerned  for  the  present  with  functions  of  one 
single  variable,  as  illustrated  by  (1)  above.  Here,  a;  is  called 
the  independent  variable,  since  we  assign  to  it  any  value  we 
like.  The  value  of  the  function,  or  more  briefly,  the  function, 
is  called  the  dependent  variable,  and  is  often  denoted  by  a 
single  letter,  as  »—/(,) 

or  y  =  2x*  —  3x  +  1. 


Fig.  1 


Graphs.     A  function 
of  a  single  variable, 

can  be  represented 
geometrical!}  by  its 
graph,  and  this  repre- 
sentation is  of  great  aid 
in  studying  the  proper- 
of  the  function. 
The  independent  vari- 
able is  laid  off  as  the 
SB-coordinate,  or  ab- 
scissa, and  the  depend 
ent    variable,   or   func- 


*  To  distinguish  between/(x)  and  F(x).  read  the  first  "small/  of  x'' 

;in<l  the  second,  "large  /'"!  s. 


INTRODUCTION  3 

tion,  as  the  y-coordinate,  or  ordinate.     Thus  the  graph  of  the 

function  ,,  .         , 

f(x)  =  z3 

is  the  curve 


,■■'■ 


Illustrations  from  Geometry  and  Physics.  The  familiar 
formulas  of  geometry  and  physics  afford  simple  examples  of 
functions.  Thus  the  area,  A,  of  a  circle  is  given  by  the 
formula 

A  =  7lT2, 

where  r  denotes  the  radius,  -k  being  the  fixed  number  3.1416. 
Here,  r  is  thought  of  as  the  independent  variable,  —  it  may 
have  any  positive  value  whatever,  —  and  A  is  the  function,  or 
dependent  variable. 

Again,  for  the  three  round  bodies,  the  volumes  are  : 

(a)  V  —  f  ttt3,  sphere  ; 

(b)  V=irr2h,  cylinder; 

(c)  V=-r%  cone. 

o 

In  (6)  and  (c),  h  denotes  the  altitude  and  r,  the  radius  of 
the  base ;  V  is  here  a  function  of  the  two  independent 
variables,  r  and  h. 

The  surfaces  of  these  bodies  are  given  by  the  formulas  : 

(a)  S  =  4vrr2,  sphere ; 

(j8)  S  =  27rr/i,  cylinder ; 

(y)  S  =  -n-rl,  cone ; 

I,  in  the  last  formula,  denoting  the  slant  height.  Thus  we 
have  three  further  examples  of  functions  of  one  or  of  two 
variables. 

The  formula  for  a  freely  falling  body  is 

where  s  denotes  the  distance  fallen  and  t  the  time ;  g  is  a 
constant,  for  it  has  just  one  value  after  the  units  of  time  and 


CALCULUS 


length  have  been  chosen.  Here,  t  is  the  independent  variable 
and  s  is  the  function.  If,  however,  we  solve  this  equation 
for  t  : 


t 

9 


then  s  becomes  the  independent  variable  and  t,  the  function. 
Sometimes  two  variables  are  connected  by  an  equation,  as 

pv  =  c, 

where  p  denotes  the  pressure  of  a  gas  and  v  its  volume,  the 
temperature  remaining  constant.  Here,  either  variable  can 
be  chosen  as  the  independent  variable,  and  when  the  equation 
is  solved  for  the  other  variable,  the  latter  becomes  the  de- 
pendent variable,  or  function.     Thus,  if  we  write 

c 

p  is  the  independent  variable,  and  v  is  expressed  as  a  function 

of  i).     But  if  we  write 

c 
p=    . 

V 

the  roles  are  reversed. 

Tlie  Independent  Variable  Restricted.  Often  the  independent 
variable  is  restricted  to  a  certain  interval,  as  in  the  cas<   of  the 

function  

y  =  Va2  —  x2. 

Here,  x  must  lie  between  —  a  and  a: 

—  a  ^  x  ^  a, 

since  other  values  of  x  make  a2  —  x~  negative,  and  the  above 
expression  has  no  meaning. 

This  was  also  the  case  with  the  geometric  examples  above 
cited.  There,  r,  h,  I  were  necessarily  positive,  since  there  is 
no  such  thing,  for  example,  as  a  sphere  of  zero  or  negative 
radius. 

The  independent  variable  may  also  be  restricted  to  being  a 
positive  whole  number,  as  in  the  case  of  the  sum  of  the  tirst  n 


v 


INTRODUCTION  5 

terms  of  a  geometric  progression : 

sn  =  a  +  ar+  ar2  -f-  ••■  +  ar71'1. 

Here, 

a  —  arn 

s„  = 

1-r 

Suppose  a  =  1,  r  =  \,  the  progression  thus  becoming 

^  2     22  2""1 

Then 

1-1 

s— 5-2-    * 

2 

and  we  have  an  example  of  a  function  with  the  independent 
variable  a  natural  number,  i.e.  a  positive  integer. 

In  the  case  of  the  functions  treated  in  the  calculus,  the  do- 
main of  the  independent  variable  is  a  continuum,  i.e.,  for  func- 
tions of  a  single  variable,  an  interval,  as 

a  ^Lx^b,  or  0  <  x. 

Ordinarily,  the  later  letters  of  the  alphabet,  particularly 
x,  y,  z,  are  used  to  represent  variables,  the  early  letters  denot- 
ing constants.  Thus  it  will  be  understood,  when  such  an  ex- 
pression as 

ax2  +  ox  +  c 

is  written  down,  that  a,  b,  c  are  constants  and  x  is  the  variable. 
Multiple-Valued  Functions;  Principal  Value.  The  expres- 
sions above  cited  are  all  examples  of  single-valued  functions ; 
i.e.  to  each  value  of  the  independent  variable  x  corresponds 
but  one  value  of  the  function.  A  function  may,  however,  be 
multiple-valued;  as  in  the  case  of  the  function  y  defined  by 
the  equation 

x2  +  y2  =  a2. 
Here 

y  =  ±  Va2  —  x2, 


6 


CALCULUS 


and  so  is  a  double-valued  function.  This  function  is,  however, 
completely  represented  by  means  of  the  two  single-valued 
functions, 

y  =  Va2  —  x2         and 


Graph  i 
x'U  y" 

>/;/.  when 

a'- 

1      x 

I             ° 

x=\a 

Fig.  2 


They  form  the  branches  of  this  multiple-valued  function. 

The  student  should  notice  that  the  radical  sign  -y/  is  defined 
as  meaning  the  positive  square  root,  not  either  the  positive  or 
the  negative  square  root  at  pleasure.  If  it  is  desired  to  ex- 
press the  negative  square  root,  the  minus  sign  must  be  written 
in  front  of  the  radical  sign,  —  ^/.  Thus  V4  =  2,  and  not  —  2. 
This  does  not  mean  that  4  has  only  one  square  root.  It  means 
that  the  notation  V4  calls  for  the  positive,  and  not  for  the 
negative,  of  these  two  roots. 

Again,  

V(-2)'=2, 

and  not  —  2.     For  (—  2)2=  4,  and  ^/  means  the  positive  root. 
And,  generally, 

J  V#2  =  x,  if  x  is  positive  ; 


(1) 


^  /-Z 


{  Vx2  =  - 


if  x  is  negative. 


A  similar  remark  applies  to  the  symbol  2y,  which  is  like- 
wise used  to  mean  the  positive  2»th  root.     Moreover, 


a2  =  Va, 


2n/— 

=  va. 


The  function 


Va 


INTRODUCTION  7 

is  often  called  the  principal  value  of  the  double-valued  function 
denned  by  the  equation 

y2  =  x. 

Since  multiple-valued  functions  are  studied  by  means  of 
single-valued  functions,  it  will  be  understood  henceforth,  un- 
less the  contrary  is  explicitly  stated,  that  the  word  function 
means  single-valued  function. 

Absolute  Value.  It  is  frequently  desirable  to  use  merely 
the  numerical,  or  absolute  value  of  a  quantity,  and  to  have  a 
notation  for  the  same.  The  notation  is:  |x|,  read  "absolute 
value  of  a?."     Thus 

j  —  3  j  =  3  and  |3|=3. 

We  can  now  write  in  a  single  formula  what  was  formerly 
stated  by  the  two  equations  (1),  namely  the  definition  of  the 
radical  sign,  ^/ : 

(2)  Vtf=|<4 

Again,  by  the  difference  of  two  numbers  we  often/  mean  the 
value  of  the  larger  less  the  smaller.     Thus  the  difference  of  4 . 
and  10  is  6 ;  and  the  difference  of  10  and  4  is  also  6.     The 
difference  of  a  and  b,  in  this  sense,  can  be  expressed  as  either 

\b—  a\  or  \a  —  b\.r 

Continuous  Functions.  A  function,  f(x),  is  said  to  be  con- 
tinuous if  a  slight  change  in  x  produces  but  a  slight  change  in 
the  value  of  the  function.  Thus  the  polynomials  ar@  readily 
shown  to  be  continuous ;  cf.  Chap.  II,  §  5,  and  all  the  func- 
tions with  which  we  shall  have  to  deal  are  continuous,  save  at 
exceptional  points. 

As  an  example  of  a  function  which  is  discontinuous  at  a 
certain  point  may  be  cited  the  function  (see  Fig.  3) 

x 


8 


CALCULUS 


When  x  approaches  the  value  0,  the  function  increases  nu- 
merically without  limit.  The  graph  of  the  function  has  the 
axis  of  y  as  an  asymptote. 

The  fractional  rational  functions  are  continuous  except  at 
the  points  at  which  the  denominator  vanishes. 

Thus  the  function 


/(*)  = 


3?2  +  l 

X'  -  1 


y  = 


is  continuous  except  at  the  points 
x  =  1  and  x  =  —  1.  Here,  the 
function  becomes  infinite.  Its 
graph  is  the  curve 

e2  +  l 
(x-l)(x  +  lY 

which  evidently  has  the  lines  x  =  1 
and  x  =  —  1  as  asymptotes. 
The  function 

Fig.  3  f(x)  =  tan  x 

is  continuous  except  when  x  is  an  odd  multiple  of  ir/2, 


x  = 


2ra  +  l 


EXERCISES 

1.  If  f(x)=x*  —  4a? +  3, 

show  that     /(l)  =  0,  /(2)  =  -  1,  /(3)  =  0. 

Compute /(0), /(I).     Plot  the  graph  of  the  function. 

2.  If  4>(x)  =  Axi 
compute  <f>(2)  and  <£(V3). 


3.    If 


F(x)  = 


2x-B 


x  +  7  ' 
compute  F(y/2)  correct  to  three  significant  figures. 


Ans.  -.0204. 


INTRODUCTION  9 

4.  If  $(#)  =  (x3  —  a;)  sin  x, 
find  all  the  values  of  x  for  which 

<f>(0)=0. 

5.  If  ij/(x)=a$  —  x~§t 
find  ^(8). 

$.    Solve  the  equation 

cc3  —  xy  +  3  =  5  y 
for  y,  thus  expressing  y  as  a  function  of  x. 

7.  If  /(*)=«*, 
show  that                    /OO/O)  =/(«  +  2/)- 

8.  It  y  = ! , 

'y      2a- 3' 

express  a;  as  a  function  of  y. 

9.  Draw  the  graph  of  the  function 

f(x)=x2  +  4:X  +  3, 

taking  1  cm.  as  the  unit. 

Suggestion  :  Write  the  function  in  the  form,  (x  -(-  1)(&  +  3). 

10.  Draw  the  graph  of  the  function 

f(x)  =  x3  —  4  x. 

11.  Draw  the  graph  of  the  function 


x1  —  4 


and  hence  illustrate  the  two  discontinuities  which  this  func- 
tion has. 

12.   Draw  the  graph  of  the  function 

i  1 

/(«J)=i- 


a;2      (a;  -  l)3 


10  CALCULUS 

13.  For  what  values  of  x  are  the  following  functions  dis- 
continuous ? 

(a)  f(x)  =  cot  x ;  (c)  f(x)  =  esc  x ; 

(b)  f(x)  =  secx ;  (d)  f(x)  =  tan  |- 

14.  Express  the  double-valued  function  defined  by  the 
equation  aj2  —  w2  =  —  1 

in  terms  of  two  single-valued  functions. 

15.  Express  the  quadruple- valued  function  defined  by  the 
equation  ?/4  _  2y2  +  x,  =  Q 

in  terms  of  four  single-valued  functions. 

16.  Express  the  sum  s„  of  the  first  n  terms  of  the  arithmetic 
progression 

a+(a  +  b)  +  (a  +  2b)+  ■••  +(a  +  n  -  16) 

as  a  function  of  n. 

Thus  obtain  the  sum  of  the  first  n  positive  integers  as  a 
function  of  ?t. 

17.  If  P  dollars  are  put  at  simple  interest  for  one  year  at 
r  per  cent,  (a)  express  the  amount  A  (principal  and  interest) 
as  a  function  of  P  and  r.  (b)  Express  the  amount  A  at  the 
end  of  n  years,  the  interest  being  compounded  annually,  as  a 
function  of  P,  r,  and  n.  (c)  Express  the  amount  A  at  the 
end  of  one  year,  if  the  interest  is  compounded  m  times  in  the 
year  at  equal  intervals,  as  a  function  of  P,  r,  m. 

2.  Continuation.  General  Definition  of  a  Function.  The  con- 
ception of  the  function  is  broader  than  that  of  the  mathemati- 
cal formulas  mentioned  in  the  last  paragraph.  Let  us  state 
the  definition  in  its  most  general  form. 

Df.finition  of  a  Function.  Hie  variable  y  is  said  to  be  a 
function  of  the  variable  x  if  there  exists  a  laio  ivhereby,  when  x 
is  given,  y  is  determined. 


INTRODUCTION  11 

Consider,  for  example,  a  quantity  of  gas  confined  in  a  cham- 
ber, —  for  instance,  the  charge  of  the  mixture  of  gasolene  and 
air  as  it  is  being  compressed  in  the  cylinder  of  an  automobile. 
The  charge  exerts  at  each  instant  a  definite  pressure,  p,  of  so 
many  pounds  per  square  inch  on  the  walls  of  the  chamber, 
and  this  pressure  varies  with  the  volume,  u,  occupied  by  the 
charge.  In  the  small  fraction  of  a  second  under  consideration, 
presumably  but  little  heat  is  gained  or  lost  through  the  walls 
of  the  chamber,  and  thus  p  is  a  function  of  v, 

In  this  case,  the  function  is  given  approximately  by  the  math- 
J  ematical  formula 

vrA 

where  C  denotes  a  certain  constant.  But  that  which  is  of  first 
importance  for  our  conception  is  not  the  formula,  but  the  fact 
that  to  each  value  of  v  there  corresponds  a  definite  value  of  p. 
In  other  words,  there  is  a  definite  graph  of  the  relation  be- 
tween v  and  p.  The  representation  of  the  relation  by  a  math- 
ematical formula  is,  indeed,  important;  but  what  we  must 
first  see  clearly  is  the  fact  that  there  is  a  definite  relation  to 
express. 

As  another  illustration  take  the  curve  traced  out  by  the 
pen  of  a  self-registering  thermometer  of  the.  kind  used  at  a 
meteorological  station.     The  instrument  consists  of  a  cylindri- 
cal drum  turned 
slowly  by  clock- 
work at  uniform 
speed     about    a 
vertical    axis,    a 

sheet    of    paper  FlG  4 

being         wound 

firmly  round  the  drum.  A  pen  is  held  against  the  paper,  and 
the  height  of  the  pen  above  a  certain  level  is  proportional  to 
the  height  of  the  temperature  above  the  temperature  corre- 


12  CALCULUS 

sponding  to  that  level.  The  apparatus  is  set  in  operation,  and 
when  the  drum  has  been  turning  for  a  day,  the  paper  is  taken 
off  and  spread  out  flat.  Thus  we  have  before  us  the  graph  of 
the  temperature  for  the  day  in  question,  the  independent 
variable  being  the  time  (measured  in  hours  from  midnight) 
and  the  dependent  variable  being  the  temperature,  represented 
by  the  other  coordinate  of  a  point  on  the  curve. 

One  more  illustration,  —  that  of  the  resistance  of  the  atmos- 
phere to  a  rifle  bullet.  This  resistance,  measured  in  pounds, 
depends  on  the  velocity  of  the  bullet,  and  it  is  a  matter  of 
physical  experiment  to  determine  the  law.  But  that  which 
is  of  first  importance  for  our  conceptions  is  the  fact  that  there 
is  a  laiv,  whereby,  when  the  velocity,  v,  is  given  an  arbitrary 
value  within  the  limits  of  the  velocities  considered,  there  cor- 
responds to  this  v  a  definite  value,  M,  of  the  resistance.  We 
say,  then,  that  R  is  &  function  of  v  and  write 

R=f(y). 

In  this  connection,  cf.  the  chapter  on  Mechanics,  §  7,  Graph 
of  the  Kesistance,  in  the  author's  Differential  and  Integral 
Calculus, 


CHAPTER    II 


DIFFERENTIATION   OF   ALGEBRAIC   FUNCTIONS 
GENERAL   THEOREMS 


1.  Definition  of  the  Derivative.  The  Calculus  deals  with 
varying  quantity.  If  y  is  a  function  of  x,  then  x  is  thought 
of,  not  as  having  one  or  another  special  value,  but  as  flowing 
or  growing,  just  as  we  think  of  time  or  of  the  expanding  cir- 
cular ripples  made  by  a  stone  dropped  into  a  placid  pond. 
And  y  varies  with  x,  sometimes  increasing,  sometimes  decreas- 
ing. Now  if  we  consider  the  change  in  x  for  a  short  interval, 
say  from  x  =  x0  to  x  =  x',  the  corresponding  change  in  y,  as  y 
goes  from  y0  to  y',  will  be  in  general  almost  proportional  to 
the  change  in  x.     For  the  ratio  of  these  changes  is 

V'  ~  .Vo 

and  this    quantity  changes    only    slightly   when   x'  is  nearly 
equal  to  x0.     Let  us  study  this_jast  statement  minutely. 


Fig.  5 


The  above  ratio  has  a  simple  geometric  meaning,  if  we  draw 
the  graph  of  the  function ;  for 

PM=x'-xQ;  MP'  =  y'-y0, 

13 


14  CALCULUS 

and  y'-fr-tanr', 

x'  -  x0 

where  r'  denotes  the  angle  which  the  secant  PP  makes  with 
the  axis  of  x.  Now  let  x'  approach  x0  as  its  limit.  Then  r' 
approaches  as  its  limit  the  angle  t  which  the  tangent  line  of 
the  graph  at  P  makes  with  the  axis  of  x,  and  hence 

1  i  in  •!-  ~  -'°  =  tan  t 
x'±x0X   —  xQ 

{  n-ad  :  "  limit,  as  x'  approaches  x0,  of  •  ,  ~  •  °    V 
V  -r  ~xo  J 

Tlie  determination  of  this  limit  and  tin  discussimhQfJtsjjiemL- 
ing  is  the  fundamentid  problem  of  the  Dtj}>  o  ntial  Calculus^ 

Such  are  the  concepts  which  underlie  the  idea  of  the  deriva- 
tive of  a  function.     Aye  turn  now  to  a  precise  formulation  of 

the  definition.     Let 
*t*.f 

(1)  n  =  /(x) 

he  a  given  function  of  x.  Xet  x0  be  an  arbitrary  value  of  x, 
and  let  y0  be  the  corresponding  value  of  the  function : 

(2)  >/o  =/(•* 

Give  to  x  an  increment,*  Ax ;  i.e.  let  x  have  a  new  value,  x', 
and  denote  the  change  in  x,  namely,  x'  —  Xq,  by  Ax : 

.(•'  —  .(■„  =  Ax,  x'  =  x0  +  Ax. 

The  function.  >/.  will  thereby  have  changed  to  the  value 

(3)  .'/'=/(•''') 

and  hence  have  received  an  increment,  Ay,  where 
.'/'  -  .'/o  =  A//,  ?/'  =  ?/0  +  A#. 

*The  student  must  not  think  of  this  symbol  as  meaning  A  times  x. 
We  might  have  used  a  single  letter,  as  h,  to  represent  the  difference  in 
question  :  x'=  Xo  +  h  ;  but  h  would  not  have  reminded  us  that  it  is  the 
increment  of  x.  and  not  of  y,  with  which  we  are  concerned.  The  notation 
is  read  "delta  x." 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS       15 


Equation  (3)  is  equivalent  to  the  following : 

(4)  ?/0  +  Ay=/(x0  +  Ax). 

From  equations  (2)  and  (4)  we  obtain  by  subtraction  the 
equation 

and  henee 

Ay  =/(x0  +  Ax)  -/(x0) 


ty  =/(zo  +  Ax)  -/(x0), 


Ax 


Ax 


Definition  of  a  Derivative.      The  limit  which   the   ratio 
(5),  namely  — ,  approaches  when  Ax  approaches  zero: 


(6) 


lim— v 

Ax±0  AX 


or 


lim/Oo  +  Ax)-/(x0) 
Ax=mj  Ax 


z's  called  the  derivative  of  y  with  respect  to  x  and  is  denoted  by 
D£y  or  DJ(x) '(read :  "Dxofy  ") : 

lim^- 


(7) 


Az=MJ  AX 


^- 


In  this  definition  Ax  may  be  negative  as  well  as  positive,  and 
the  limit  (6)  must  be  the  same  when  Ax  approaches  0  from  the 
negative  side  as  when  it  approaches  0  from  the  positive  side. 

To  differentiate  jl 
function  is  to  tinrl  its 
derivative^ 

The  geometrical  in- 
terpretation-, of  the 
analytical  process  of 
differentiation  is  to 
find  Jthe  slope  ©£-the- 
graph  ofjtheXunction. 

tan  t'  =  — " 
Ax 

and 


y 

pa/ 

Z^L 

y\fy 

°^. ^r      \&X 

i 
i 

yd 

Y 

i 

i 

X 

0 

*v 

X' 

Fig.  6 


tan  t  =  lim  tan  r'  =  lim  — "  =  Dxy. 

y=p  Ai^o  Ax 


16  CALCULUS 

2.  Differentiation  of  xn.  Suppose  n  has  the  value  3,  so  that 
it  is  required  to  differentiate  the  function 

(1)  y  =  x>. 

We  must  follow  the  definition  of  §  1  step  by  step.  Begin, 
then,  by  assigning  to  x  a  particular  value,  x0,  which  is  to  be 
held  fast  during  the  rest  of  the  process,  and  compute  from 
equation  (1)  the  corresponding  value  y0  of  y  : 

(2)  2/o  =  *o3- 

Next,  give  to  x  an  arbitrary  increment,  Ax,  denote  the  corre- 
sponding increment  in  y  by  Ay,  and  compute  it.  To  this  end 
we  first  write  down  the  equation 

(3)  2/o  +  Ay=(.r0+Ax-)3. 

The  right-hand  side  of  this  equation  can  be  expanded  by  the 
binomial  theorem,  and  hence  (3)  can  be  written  in  a  new  form  :* 

(4)  3/0  +  Ay  =  x03  +  3  x<?Ax  +  3  x*Ax*  +  Ax-3. 

Subtract  equation  (2)  from  equation  (4) : 

Ay  =3  Xq-Ax  +  3  xQAx"-  +  Ax3. 

Next,  divide  through  by  Ax: 

^l=3x02  +  3  x0Ax  +  Ax  . 
Ax 

We  are  now  ready  to  let  Ax  approach  0  as  its  limit : 
lim  *y  _  lim (3 x  2  +  3  x  \x  +  Ax2). 

*It  is  at  this  point  that  the  specific  properties  of  the  function  x3  conie 
into  play.  Here,  it  is  the" binomial  theorem  that  enables  us  ultimately 
to  compute  the  limit.  In  the  differentiations  of  later  paragraphs  and 
chapters  it  will  always  be  some  characteristic  property  of  the  function  in 
hand  which  will  make  possible  a  transformation  at  this  point. 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS       17 

The  limit  of  the  left-hand  side  is,  by  definition,  Dxy.     On  the 

right-hand  side,  each  of  the  last  two  terms  in  the  parenthesis 

approaches  the  limit  0,  and  so  their  sum  approaches  0,  also. 

The  first  term  does  not  change  with  Ax.     Hence,  the  whole 

parenthesis  approaches  the  limit  3x02.     We  have,  then,  as  the 

final  result :  ~  0    „ 

Dxy=3x02. 

The  subscript  has  now  served  its  purpose,  which  was,  to 
remind  us  that  x0  is  not  to  vary  with  Ax,  and  it  may  be 
dropped.     Thus  Dxx*  =3x*. 

The  differentiation  of  the  function  xn  in  the  general  case 
that  n  is  any  positive  integer  can  be  carried  through  in  pre- 
cisely the  same  manner.     As  the  result  of  the  first  step  we  have 

(5)  y0  =  x0\ 
Next  comes : 

(6)  y0  +  Ay  =  («o  +  Ax)", 

and  we  now  apply  the  binomial  theorem  to  the  expression  on 
the  right-hand  side.     Thus 

(7)  y0  +  Ay  =  x0n  +  nx^Ax  +w<>~    )  3;0"~2A.t2  +  •••  +  Ax\ 
On  subtracting  (5)  from  (7)  we  have  : 

Ay  =  ?ix0n~1Ax  -| — i- — — -J- x0n  -Ax-  +  —  +  &®  • 

Now  divide  through  by  Ax : 

A?/  „_,   ,  n(n  —  T)     „_„A      ,  ,    A   __i 

-^  =  nxa"  x  +    \         ^o    *Ax  +  ■■■  +  Axn  \ 

Ax  1-2 

and  let  Ax  approach  the  limit  zero : 

lim  ^  =  lim  (way-1  +  n^n  ~  ^  x^Ax  +  •  •  •  +  Aa5»-*\ 
Aa^o  A#     Alio  \  1-2  y 

Each  term  of  the  parenthesis  after  the  first  is  the  product 
of  a  constant  factor  and  a  positive  power  of  Ax.     This  second 


18  CALCULUS 

factor  approaches  zero  when  Ax  approaches  zero  ;  consequently 
the  whole  term  approaches  zero.  There  is  only  a  fixed  num- 
ber of  these  terms,  and  so  the  whole  parenthesis  approaches 
the  limit  nx0n~\     Hence  ' 

Dxy  =  nx0"-\ 

On  dropping  the  subscript  we  obtain  the  final  result 

(8)  I)xxn  =  nxn~l. 
In  particular,  if  n  =  1,  we  have 

(9)  Dxx  =  l. 

EXERCISES 

Differentiate  the  following  seven   functions,  applying  the 
process  of  §  1  step  by  step. 

1.  y  =  \  .r'\  Ans.    Dxy  =  \2x2. 

2.  y  =  or4. 

v£.  ?/  =  2  x:  —  8  x  +  1.           An*.    I)  y  =  \  x  —  3. 

4.  y  =  x1  —  x5.                  Ans.    D.y  =  7  x&  —  5x*. 

5.*  f(x)=l-2x*.                 Ans.   DJ(x)=-Sx3. 

6.  <f>  (.*•)  =  x*  —  2  a  +  1. 

7.  F(x)  =  (l-.r):  ay-  % 

8.  Let  y  =  5x  —  xl, 

and  take  x0  =  1 ;  then  y0  =  4.     If  Ax  =  .2,  then  Ay  =  .56  and 

^  =  2.8.     Show  further  that, 
Ax 

for    Aaj  =  .l,  Ay  =  .29,  ^=2.9; 

Ax 
and 

for    Ax  =  M,  Ay  =  .0299,  ^  =  2.99. 

Ax 

*  It  is  immaterial  whether  we  write 

?/  =  l-2x4  or  /(x)=l-2x4. 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS       19 


Plot  the  curve  accurately  for  values  of  x  from  x  =  0  to 
x  =  5,  taking  1  cm.  as  the  unit,  and  draw  the  secants  *  in 
each  of  the  three  foregoing  cases. 

What  appears  to  be  the  slope  of  the  curve  at  the  point 
(x0,  y0)  =  (l,  4)?     Prove  your  guess  to  be  correct. 

9.   In  Ex.  7,  let  x0  =  —  1.     If  A&  is  given  successively  the 

values  .01  and  —  .01,  compute  Ay  and  — ^« 

Ax 

10.    Complete  the  following  table  : 


Aa: 

Ay 

tanr'  =  ^ 
Ax 

.1 

.01 

.001 

for  each  of  the  functions  : 

(a)  y  =  x1-  -  2x  +  1, 

(b)  y  =  x-x3, 
{&  y  =  3xi  —  x, 


x0  =  2; 

xo  —  — ~  1 ; 
xQ  =  0. 


11.    By  means  of  the  general  theorem  (8)  write  down  the 
derivatives  of  the  following  functions  : 

By  means  of  the  definition  of  §  1  differentiate  each  of  the 
following  functions  : 


13. 
J4. 


y  = 


H<^ 


i 

Xs 


Ans.    Dj/  =  —  — 


Aus.    Dzy  =  — 


Ans.   Dry  = • 

.T4 


*  The  student  should  recall  from  his  earlier  work  how  to  draw  a  straight 
line  on  squared  paper  when  a  point  and  the  slope  of  the  line  are  given. 


20  CALCULUS 

3.   Derivative  of  a  Constant.     The  function  i 

f{x)=c, 

where  c  denotes  a  constant,  has  for  its  graph  a  right  line  paral- 
lel to  the  axis  of  x.  Since  the  derivative  of  a  function  is  repre- 
sented geometrically  by  the  slope  of  its  graph,  it  is  clear  that 
the  derivative  of  this  function  is  zero  : 

Drc  =  0. 

It  is  instructive,  however,  to  obtain  this  result  analytically 
by  the  process  of  §  1.     We  have  here : 

.Vo=/Oo)=c, 

y0  +  Ay  =  f(x0  +  Ax)  =  c  ; 

hence  Ay  =  0         and  ^=0. 

Ax 

Now  allow  Ax  to  approach  0.  The  value  of  Ay/ Ax  is  always 
0,  and  hence  its  limit  *  is  0 : 

lim  ^L  =  0,  or  Dxc  =  0. 

aasM)  Aa; 

*  We  note  here  an  error  frequently  made  in  presenting  the  subject  of 
limits  in  school  mathematics.  It  is  there  often  stated  that  "  a  variable  X 
approaches  a  limit  A  if  X  comes  indefinitely  near  to  A,  but  never  reaches 
A."  This  last  requirement  is  not  a  part  of  the  conception  of  a  variable's 
approaching  a  limit.  It  is  true  that  it  is  often  inexpedient  to  allow  the 
independent  variable  to  reach  its  limit.  Thus,  in  differentiating  a  func- 
tion, the  ratio  Ay/ Ax  ceases  to  have  a  meaning  when  Ax  =  0,  since  divi- 
sion by  0  is  impossible.  The  problem  of  differentiation  is  not  to  find  the 
value  of  Ay/ Ax  when  Ax  =  0  ;  such  a  question  would  be  absurd.  What 
we  do  is  to  allow  Ax  to  approach  zero  as  its  limit  without  ever  reaching 
that  limit.  We  can  do  this  for  the  reason  that  Ax  is  the  independent 
variable. 

When,  however,  it  is  Ay  or  Ay/ Ax  that  is  under  consideration,  we  have 
to  do  with  dependent  variables,  and  we  have  no  control  over  them,  as  to 
whether  they  reach  their  limit  or  not.  Thus  in  the  case  of  the  text  both 
Ay  and  Ay/Ax  are  constants  (=0).  When  Ax  approaches  0,  they  always 
have  one  and  the  same  value,  and  so,  under  the  correct  conception  of 
approach  to  a  limit  each  approaches  a  limit,  namely  0. 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS       21 

We  can  state  the  result  by  saying :   The  derivative  of  a  con- 
stant is  0. 

4.   Differentiation  of  y/x.     Let  us  differentiate 
*/=  Vs. 
Here,  y0  =  V»0>  y0  +  Ay  =  Vx0  +  Ax, 


Ay  __  -\/xQ  -f-  Ax  —  Vxq 

A.*-  A.c 

We  cannot  as  yet  see  what  limit  the  right-hand  side  approaches 
when  Ax  approaches  0,  for  both  numerator  and  denominator 

approach  0,  and  -  has  no  meaning.  We  can,  however,  trans- 
form the  fraction  by  multiplying  numerator  and  denominator 
by  the  sum  of  the  radicals  and  recalling  the  formula  of  Elemen- 
tary Algebra  :  ai  _  &2  =(a  _  &)(a  +  &) 


Thus   Ay __  V^o  +  Ax—^/x0  _  Vflp  +  Ax+VxQ 
Ax-                  Ax                  Vav,  +  Aaj+V.fo 
_   1  (x0  -}-  Aa)  —x0_ 


A.c      Vft0  +  Ax-  +  -\A-0      Vft0  +  Ax  +  Vx0' 

and  hence    lim  — ^  =  lim  — ==== —  = —  • 

ax^o  Ax     a*mj  V^  +  Ax  +  Vx0     2  Vx0 

Dropping  the  subscript,  we  have  : 


2Vx- 

EXERCISES 

-/  1  1 

1.   Differentiate  the  function  y  = ^4ws.  Z).j/  = —  ■ 

Vx  2y/x* 


2.    If  y=V2-3a;, 

—  3 
show  that  Dxy  = ■  • 

4  2V2-3x 


22  CALCULUS 


3.    Prove:  DxVl  —  x  = 


4.    Prove  :  Dz^/a  +  bx  = 


2Vl-x 
b 


2  Va  +  bx 


5.  Three  Theorems  about  Limits.  Infinity.*  In  the  further 
treatment  of  differentiation  the  following  theorems  are  needed. 

Theorem  I.  Tlie  limit  of  the  sum  of  two  variables  is  equal  to 
the  sum  of  their  limits: 

lim  (X  +  Y)=  lim  X  +  lim  V. 

In  this  theorem  we  think  of  X  and  Y  as  two  dependent 
variables,  each  of  which  approaches  a  limit : 

limX=A,  lim  T=B. 

We  do  not  care  what  the  independent  variable  may  be.     In. 
the  applications  of  the  theorem  to  computing  derivatives,  the 
independent  variable  will  always  be  Ax,  and  it  will  be  allowed 
to  approach  0,  without  ever  reaching  its  limit. 

Since  X  approaches  A,  it  comes  nearer  and  nearer  to  this 
value.  Let  the  difference  between  the  variable  and  its  limit 
be  denoted  by  e  ;  then  the  limit  of  e  is  0  : 

(1)  X-A  =  e}  X=A  +  e; 

lim  £  =  0. 

Similarly,  let 

(2)  .  Y-B  =  v,  Y=B  +  V; 
then                                          lim  77  =  0. 

*  This  paragraph  should  be  read  carefully  and  its  content  grasped, 
tint  the  student  .should  not  be  required  to  reproduce  it  at  this  stage  of  his 
work.  He  will  meet  frequent  applications  of  its  principles,  and  he  should 
turn  back  each  time  to  these  pages  ami  read  anew  the  theorem  involved. 
with  its  proof.  When  he  has  thus  come  to  see  the  full  meaning  and  im- 
portance of  these  theorems,  he  should  demand  of  himself  that  he  be  able 
readily  to  reproduce  the  proofs. 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS       23 

It  will  be  convenient  to  think  of  these  numbers  as  repre- 
sented geometrically  by  points  on  the  scale  of  numbers,  thus  : 

1 1 1 1 1 1 

0  1  AX  Y     B 

Fig.  7 

Of  course,  A  and  B  may  be  negative  or  0.     c  and  rj  may  be 
negative  as  well  as  positive,  or  even  0. 

Consider  the  variable  X  +  Y.     Its  value  from  (1)  and  (2)  is  : 

X  +  Y=A  +B+.e  +  v. 

Hence  lim  (X  +  Y)  =  lim  (A  +  B  -f  e  +  rj). 

But  since  lim  e  =  0  and  lim  rj  =  0,  the  limit  of  the  right-hand 
side  of  this  equation  is  A  -+-  B,  or 

\im(X  +  Y)=A  +  B. 

Consequently,     lim  (X  +  7)=  lim  X  -f  lim  Y,  q.  e.  d. 

Corollary.  The  limit  of  the  sum  of  any  fixed  number  of 
variables  is  equal  to  the  sum  of  the  limits  of  these  variables : 

lim  (X]  +  Xo  +  •••  +  X„)  =  lim  Xl  +  lim  X2  +  •»  -f-  lim  X„. 

Suppose  n  —  3.     Then 

Xj  +  X2  +  X3  =(X,  4-  X2)+  X3. 

From  Theorem  I  it  follows  that 

lim  (Xx  +  Xo  +  X3)  =  lim  (Xt  +-X2)  +  lim  X3- 

Applying  the  Theorem  again,  we  have 

lim  (Xx  +  X2)  =  lim  Xx  +  lim  X2. 

Hence  the  corollary  is  true  for  n  =  3.  It  can  now  be  estab- 
lished for  n  =  4  ;  and  so  on.  By  the  method  of  Mathematical 
Induction  it  can  be  proven  generally.  Or,  the  proof  of  the 
main  theorem  may  be  extended  directly  to  the  present 
theorem. 


24  CALCULUS 

Theorem  II.  The  limit  of  the  product  of  two  variables  is 
equal  to  the  product  of  their  limits : 

lim  (AT)  =  (lim  A')(lim  Y). 
From  equations  (1)  and  (2)  it  follows  that 
XY=(A  +  e)(B  +  rj), 
or  AT-  .  I B  +  Be  +  Av  +  ev. 

Hence  lim  XY=  lim  (AB  +  Be  +  Av  +  eV). 

Since  A  and  B  are  constants,  each  of  the  last  three  terms  in 
the  parenthesis  approaches  the  limit  0,  and  so  the  limit  of  the 
parenthesis  is  AB.     Hence 

lim  (XY)=AB, 

or  lim(Xr)  =  (limA)(]iin  F  I,  q.e.d. 

Corollary.     Tlie  limit  of  the  product  of  n  variables  is  equal 

to  the  product  of  the  limits  of  these  variables: 

lim  (Xi  X2 .»  X, )  =  (lim  ^(lim  X2)  -  (lim  Xn). 
The  proof  is  similar  to  that  of  the  corollary  under  Theorem  I. 
Remark.     As  a  particular  case  under  Theorem  II  we  have  : 
lim(CX)  =  C'(limX), 
where  C  is  a  constant. 

Theorem  III.  The  limit  of  the  quotient  of  two  variables  is 
equal  to  the  quotient  of  their  limits,  provided  that  the  limit  of  the 
divisor  is  not  0 :  „     ..      „ 

hmA  =  hm  -\        if  lim  r^  a 
Y     lim  F' 

From  equations  (1)  and  (2)  above  we  have : 

A  =  A  +  e 
Y     B  +  v' 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS       25 

Subtract  A/B  from  each  side  of  this  equation  and  reduce : 

X     A=A  +  *     A=Be-Av  Jx^ 
Y      B     B  +  v      B      B2  +  Bv 

Hence  |°4  +  ^"^» 

Y     B      B2  +  Bv 

and  lim  —  =  lim(  — H — 2Y 

We  wish  to  show  that 

,  •       Be  —  Art       A 

hni '-  =  0. 

The  numerator  is  seen  at  once  to  approach  zero.     The  limit  of 
the  denominator  is  B2.     Let  H  be  a  positive  number  less  than 


0  H  B< 

Fig.  8 


B2.  Then  the  denominator  will  finally  become  and  remain 
greater  than  H,  and  hence  the  numerical  value  of  the  quotient 
in  question  will  not  exceed  the  numerical  value  of 


But  the  limit  of  this  expression  is  zero,  and  hence 

..     X     A 

Inn  —  =  — , 
Y     B 

lim|  =  ^f,  q.e.d. 

Y      hm  Y 

In  particular,  we  see  that,  if  a  variable  approaches  unity  as 
its  limit,  its  reciprocal  also  approaches  unity  : 

If  lim  X  =  1,         then        lim  -^  =  1. 

2L 


26 

CALCULUS 

Also, 

lim.  — 
X 

C 

limX' 

where  C 

is 

a 

constant 

and  lim 

A>0. 

Remark.     If  the  denominator  T  approaches  0  as  its  limit, 

no  general  inference  about  the  limit  of  the   fraction  can  be 

drawn,  as   the   following  examples    show.      Let    Y  have   the 

values  : 

111  1 


Y= 


10'   100'   1000'      '  10  ■' 


(1)  If  the  corresponding  values  of  X  are  : 
111  1 


X  = 


102'   100-'   10002'      '  102"' 


X  1 

then  lim  —  =  lim  —  =  0. 


(2)     If  X  = 


Y  10" 

111  1 


vio'  vioo'  v1000'    '  10|' 


then  X/y=  10"/2  approaches  no  limit,  but  increases  beyond 
all  limit. 

(3)     If  X=  c 


10'   100'  1000'      '  10"' 
where  c  is  any  arbitrarily  chosen  fixed  number,  then 


(4)  If  X= 


lim  —  =  c. 
Y 

111 


10'       100'   1000'       10,000' 


then  Xj  Y  assumes  alternately  the  values  +  1  and  —  1,  and 
hence,  although  remaining  finite,  approaches  no  limit. 

To  sum  up,  then,  we  see  that  when  X  and  Y  both  approach 
0  as  their  limit,  their  ratio  may  approach  any  limit  whatever, 
or  it  may  increase  beyond  all  limit,  or  finally,  although  remain- 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS       27 

ing  finite,  i.e.  always  lying  between  two  fixed  numbers,  no  mat- 
ter how  widely  the  latter  may  differ  from  each  other  in  value, 
—  it  may  jump  about  and  so  fail  to  approach  a  limit. 

Infinity.     If   lim  X  =  A  =£  0   and   lim  Y=  0,  then  X/  Y  in- 


creases beyond  all  limit,  or  becomes  infinite.  A  variable  Z  is 
said  to  become  infinite  when  it  ultimately  becomes  and  re- 
mains greater  numerically  than  any  preassigned  quantity,  how- 
ever large.*  If  it  takes  on  only  positive  values,  it  becomes 
positively  infinite;  if  only  negative  values,  it  becomes  negatively 
infinite.     We  express  its  behavior  by  the  notation : 

lim  Z  =  oo      or     lim  Z  =  +  co      or     lim  Z  =  —  go  . 

But  this  notation  does  not  imply  that  infinity  is  a  limit ;  the 
variable  in  this  case  approaches  no  limit.  And  so  the  notation 
should  not  be  read  "  Z  approaches  infinity "  or  "  Z  equals 
infinity  ;  but  "  Z  becomes  infinite." 

Thus  if  the  graph  of  a  function  has  its  tangent  at  a  certain 
point  parallel  to  the  axis  of  ordinates,  we  shall  have  for  that 
point : 

lim  — "  =  oo  ; 

Ax=M)  A# 

read  :    "  Ay/ Ax  becomes  infinite  when  Ax  approaches  0." 

Some  writers  find  it  convenient  to  use  the  expression  "a 
variable  approaches  a  limit "  to  include  the  case  that  the  vari- 
able becomes  infinite.  We  shall  not  adopt  this  mode  of  ex- 
pression, but  shall  understand  the  words  "  approaches  a  limit " 
in  their  strict  sense. 

If  a  function  f(x)  becomes  infinite  when  x  approaches  a  cer- 
tain value  a,  as  for  example 

f(x)  =  -     for     a  =  0, 

*  Note  that  the  statement  sometimes  made  that ' '  Z  becomes  greater 
than  any  assignable  quantity  "  is  absurd.  There  is  no  quantity  that  is 
greater  than  any  assignable  quantity. 


28  CALCULUS 

we  denote  this  by  writing 

/(a)  =00 

(or    /(a)=+oo      or     =—00,  if  this  happens  to  be  the  case 
and  we  wish  to  call  attention  to  the  fact). 
It  is  in  this  sense  that  the  equation 

tan  90° =00 

is  to  be  understood  in  Trigonometry.  The  equation  does  not 
mean  that  90°  has  a  tangent  and  that  the  value  of  the  latter  is 
00.  It  means  that,  as  x  approaches  90°  as  its  limit,  tana; 
exceeds  numerically  any  number  one  may  name  in  advance, 
and  stays  above  this  number  as  x  continues  to  approach  90° 
without  ever  reaching  its  limit,  90°. 

—  Definition  of  a  Continuous  Function^  ~  We  can  now  make  more 
explicit  the  definition  given  in  Chapter  I  by  saying:  f(x)  is 
continuous  at  the  point  x  =  a  if 

\imf(x)=f(a). 

From  Exercises  1-3  below  it  follows  that  the  polynomials 
are  continuous  for  all  values  of  x,  and  that  the  fractional 
rational  functions  are  continuous  except  when  the  denominator 
vanishes. 

EXERCISES 

1.  Show  that,  if  n  is  any  positive  integer, 

lim(A~")  =  (limX)\ 

2.  If  G  (x)  =  c0  +  <hx  +  c2x2  +  •"  +  c„2n> 

then  lim  G  (x)=  G  (a)=  c0  +  cxa  -f-  c^a2  -}-•••+  cnan. 

3.  If  G(x)  and  F(x)  are  any  two  polynomials  and  if  F(a)^0, 

then  lim«M=e(5}. 

*±a  F(x)      F(a) 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS       29 

4.  If  X  remains  finite  and  Y  approaches  0  as  its  limit,  show 
that  lim(XF)=0. 

5.  Show  that  ,.  x2  +  1        _  1 

xL^3x2  +  2a;-l~3' 

Suggestion.     Begin  by  dividing  the  numerator  and  the  de- 
nominator by  x2. 

Evaluate  the  following  limits : 

6.  lim  —    — 7.    lim — t —   . 

x=o0  .^3  _7  a-  +  3  x=-0  4,r6  _|_  3^4  +  7X2_  ]_ 

,.     a# -|- far1  _     v     arc  +  fta;-1 

8.    lim ! 9.    lim ■ 

x=oo  ex  -\-  dx  l  x=o  ex  +  da;-1 


10.    lim ■ 11.    hm  — 

*-°°        •'"  x==°  V3  +  5x2  +  4x4 

12.    lim- ! - 13.     hm 


Vl  +  x4 
6.   General  Formulas  of  Differentiation. 

Theorem  I.  The  derivative  of  the  product  of  a  constant  and 
a  function  is  equal  to  the  product  of  the  constant  into  the  deriva- 
tive of  the  function : 

(I)  DXcu)=cDjc. 

For,  let  y  =  cu. 

Then  y0  =  cu0, 

y0  +  &y  =  c(u0  +  Au), 

hence  Ay  =  cAu, 

Ay  _    An 

Ax        Ax 

and  lim^  =  lin/c^ 

&x±s>Ax  ■    ±x±o\   Ax 


30  CALCULUS 

The  limit  of  the  left-hand  side  is  Dxy.  On  the  right,  Au/Ax 
approaches  Dxu  as  its  limit.  Hence  by  §  5,  Theorem  II,  the 
limit  of  the  right-hand  side  is  cl)ji,  and  we  have 

Dx(cu)=cDxu,  q.e.  d. 

Theorem  II.      TJie  derivative  of  the  sum  of  two  functions  is 
equal  to  the  sum  of  their  derivatives: 


(II) 

Dx(u  +  v)=Dxu  +  Dxv. 

For, 

let 

y  =  u  +  v. 

Then 

?/o  =  "o  +  «o> 
y0  +  Ay  =  Uq  +  Aw  +  u0  +  Ai>, 

hence 

Ay  =  Au  +  Aw, 

and 

A,v  __  A"       A/' 
A/0      Ax      Ax 

When  A.r  approaches  0,  the  first  term  on  the  right  approaches 
Dxu  and  the  second  Dzv.  Hence  by  §  5,  Theorem  I,  the  whole 
right-hand  side  approaches  Dxu  +  Dxv,  and  we  have 


t     Ay      v     (An  ,  Av\      ,•     Au,  r 
Inn  ^-  =  hm 1 =  lim 1—  In 

AxioAft         Ax=0\A.T         AX  J         Az=oA£         Ai; 

or  />_//  =  l),u  +  Z)xt',  q.  e.  d. 


Ay 
im  — , 

^i=o  Aa 


Corollary.  77<e  derivative  of  the  sum  of  any  number  of 
functions  is  equal  to  th<  sum  of  their  derivatives. 

If  we  have  the  sum  of  three  functions,  we  can  write 

u  -f  v  +  w  =  u  -\-{v  +  w). 
Hence  Dx(u  +  v  +  w)  =  Dxii  +  Dx(v  +  tv) 

=  Dxu  +  Dxv  +  Z>xw. 

Next,  we  can  consider  the  sum  of  four  functions,  and  so  on. 
Or  we  can  extend  the  proof  of  Theorem  II  immediately  to  the 
sum  of  n  functions. 


DIFFERENTIATION   OF  ALGEBRAIC  FUNCTIONS       31 

Polynomials.     We  are  now  in  a  position  to  differentiate  any 
polynomial.     For  example : 

D/To4  -  Sx3  +  x  +  2) 

=  DJ1  x4)  +  Dx(-  5  a;3)  +  D£x  -f  Dx2 

=  7  Dxx*  -  5  Dxx*  +  1  =  28  x*  -  15  x2  +  1. 

EXERCISES 

Differentiate  the  following  functions  : 
l/  ?/  =  2x2  -  3x  +  1.  ^ws.    D.y  =  4a;  -  3. 

g";  y  =  a  +  bx  +  ex2.  -4ws.    2>x?/  =  &  -f-  2  ex. 

S;  y  =  ic*  —  3 x3  +  x  —  1.      Ans.    Dzy  =  4X3  —  9  x-  +  1. 

4.  ?/  =  a  +  bx  -f  ex2  -f-  dx3. 

x6  —  3x*  —  2x-f  1  .        o  b      r  ^      1 

5.  y  = ! Ans.    oar  —  ox*  —  !. 


*f      m  =  ^±^±l.  Ans. 


ax  +  b 
2/i  h 

l/  ttx4  —  3f  x2  +  V3.  Ans.   4ttx3  —  7.\ x. 

8.    Differentiate 

(a?fv0t  —  16 12  with  respect  to  £ ; 

(6)   a  -f-  &s  +  es2  with  respect  to  s  ; 

(c)  .Olfo/4  —  8.15m?/2 —  .9£m  with  respect  to  y. 

9/Find  the  slope  of  the  curve 

4y  =  x4  —  8x  —  1 
at  the  point  (1,  —  2).  Ans.    —  1. 

10.    At  what  angle  does  the  curve 
Sy  =  4x  —  x3 
cut  the  negative  axis  of  x? 


32  CALCULUS 

11.  At  what  angles  do  the  curves  y  =  x-  and  y  =  x3  intersect  ? 

Ans.   0°  and  8°  7'. 

12.  At  what  angles  do  the  curves  y  =  xz  —  3x  and  y  —  x 
intersect  ?  .  1  ns.   26°  34'  and  38°  40'.  / 

7.    General  Formulas  of  Differentiation,  Continued. 

Theorem  III.     The  derivative  of  a  product  is  given  by  the 

formula  : 

(III)  Dx(uv)=uDxv  +  vDxu. 

Let  y  =.  uv. 

Then  y0  =  u0v0, 

y0  +  A?/  =(m0  +  Aw)  (r0  +  Av), 

Ay  =  »0Av  +  UqAm  -f-  AuAv, 

Aw  Au  ,       Am    ,    .     Av 

Ax  A.c  A.i-  A.»' 

and,  by  Theorem  I,  §  5 : 

Ax^=oAa:       ai=mj\     Axy       A*=y)\     A#/      a/=Hi^       A.r 

By  Theorem  II,  §  5,  the  last  limit  has  the  value  0,  since 
bin  Aw  =  0  and  lim  (Ay/Aa;)  =  Dxv.  The  first  two  limits  have 
the  values  UqD.v  and  v0Dxii  respectively.*  Hence,  dropping 
the  subscripts,  we  have  : 

Dxy  =  uDxv  +  vl)  '/.  q.  e.  d. 

By  a  repeated  application  of  this  theorem  the  product  of 
any  number  of  functions  can  be  differentiated.     When  more 

*  More  strictly,  the  notation  should  read  here,  before  the  subscripts 
are  dropped  :  [DxUJx^,  etc.  Similarly  iu  the  proofs  of  Theorem  I,  II, 
aud  V. 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS       33 

than  two  factors  are  present,  the  formula  is  conveniently  writ- 
ten in  the  form : 

/-i\  DJuvw)      Dm  ,  DTv  .  DTw 

s      (1)  — ^ -=  ^ 1 S 1 — ' 

UVIV  U  V  IV 

For  a  reason  that  will  appear  later,  this  is  called  the  loga- 
rithmic derivative  of  uvio. 

Theorem  IV.     The  derivative  of  a  quotient  is  given  by  the 
formida  ;* 


(IV) 

Y)  fu\  —  vDxu  —  uDxv 

\v  J                   V2 

Let 

u 

y=— 

V 

Then 

y0  =  u\             y0  +  Ay  =  U»  +  *u 

v0                                v0  +  Au 

a  ?/  —  uo  +  Al(     uo  —  vo&u  —  "<A» 

v0  +  Ay      v0      v0(v0  +  Av) 

Au  Ay 

v0 u0 — 

A?/  _     Ax  Ax 

Ax~  v0(v0  +  Av) 

By  Theorem  III  of  §  5  we  have : 

v     /    A?*         Av\ 

lim   v0 ifn  — 

..     A?/      Ar^=oV     Ax  Ax  J 

hm  — -  = - -  • 

Ax^o  Ax       lim  [v0  (y0  +  Av)] 

Applying  Theorems  I  and  II  of  §  5  and  dropping  the  sub- 
scripts we  obtain : 

Dy  =  vDxu.-v,BxV)  qed 

v'1 

*  The  student  may  find  it  convenient  to  remember  this  formula  by 
putting  it  into  words:  "The  denominator  into  the  derivative  of  the 
numerator,  minus  the  numerator  into  the  derivative  of  the  denominator, 
over  the  square  of  the  denominator." 


34  CALCULUS 

Example  1.     Let  _  2  —  3a; 

V~ '  \-2x 

(l-2x)Dx(2-Sx)-(2-3x)DAl-2x) 
xnen      uzy-  (l_2x)2 

^(l-2a;)(-3)-(2-3.r)(-2)_        1 

(l-2x)2  (l-2x)2' 

Example  2.     To  prove  that  the  theorem 
Dzxn  =  BX""1 
is  true  when  n  is  a  negative  integer,  n  =  —  m.     Here 

af 

r,    „      x'"DA  —  lDxxm         mxm~l         m^m-t 

Hence     Dxx"  = ■* — —  = —  =  —  mx  m  \ 

x-m  x 

On  replacing  m  in  this  last  expression  by  its  value,  —  n,  the 
proof  is  complete. 


Ans.    Dxy- 


EXERCISES 

Differentiate  the  following  functions 

1. 

X 

y~~l-x2 

2. 

1 

y  ~  1  +  x2 

3. 

X3 

y  =  z 

1  —  X 

4. 

X2 

y      1  +  x 

5. 

1-t 

s  = 

1+t 

fi. 

z2  +  a2 

(1  -  a2)2 


o, 


Ans.    Dzy  = 


Ans.    l)sy  = 


(1  +  x*y 

?>x2-2x* 
(1  -  a)2  ' 


A         n         2x  +  x2 
Aiis.    Djj  = X — . 

J      (1  +  x)2 
Ans.    Dts  = 


a  +  ty 

ins     z2  +  2az-a-\ 
z  +  a  '    z2  +  2az  +  a2 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS       35 
7        2ay  8         ax+b 


9.    ?-^L.  10. 


a?  —  y* 

x3  +  a3 

a;  +  a 

a  +  to  +  ca'2 

a; 

3  —  4:X  +  X3 

x2  +px  -f-  ^ 
x2  +  a2 


tc*  +  a4 

x2 

a;2  a? 

8.   General  Formulas  of  Differentiation,   Concluded.     Com- 
posite Functions. 

Theorem  V.     If  v.  is  expressed  as  a  function  of  y  and  y  in 
turn  as  a  function  of  x  : 

u=f(y),  y  =  <t>(;x), 

then 

(V)  Dxu  =  Duu.Dty. 

Here  ?/o  =  </>«>,  «o=/(2/o), 

yQ  +  A?/  =  <£(a\)  +  Aa;),  uQ  +  Aw  =/Oo  +  A?/), 

A?i=/(y0  +  Ay)  -/(%), 


A^  ^/(?/o  +  A?/)  -/Q0)  #  Aj/# 
Aa;  Ay  Aa: 

When  Aa;  approaches  0,  A?/  also  approaches  0,  and  hence  the 
limit  of  the  right-hand  side  is 

(lim  f{yQ+^)-m\(lim  ajA = Df{y)DiV. 

\Ay±o  Ay  J  \Axi0  Aa:/ 

The  limit  of  the  left-hand  side  is  Dzu.     Consequently 

Dxu  =  DyM  •  Bjj,  q.  e.  d. 

This  equation  can  also  be  written  in  the  form : 
(V)  Dxu  =  DJ(y)DMx)- 


36  CALCULUS 

The  truth  of  the  theorem  does  not  depend  on  the  particular 
letters  by  which  the  variables  are  denoted.  We  may  replace, 
for  example,  x  by  t  and  y  by  x  : 

Dtu  =  Dzii  Dtx. 

Dividing  through  by  the  second  factor  on  the  right,  we  thus 
obtain  the  formula : 

(V")  D**  =  jp- 

Example!.  In  §  4  we  differentiated  the  function  V#,  and 
we  saw  that  other  radicals  can  be  differentiated  in  a  similar 
manner.  But  each  new  differentiation  required  the  evaluation 
of  lim  Ay /Ax  by  working  through  the  details  of  a  limiting 
process.  Theorem  V  enables  us  to  avoid  such  computations, 
as  the  following  example  will  show. 

To  differentiate  the  function 


u  =  Va2  —  Xs. 

Let  y  =  a2  —  x2. 

Then  u  =y/y, 

and  the  differentiation  thus  comes  directly  under  Theorem  V, 
if  we  set 

fit)  =  Vy,  <t>  O)  =  a2  -  x\ 

Hence  we  have  :     . 

(1)  Dxu  =  D„VyDx(a*-&). 

Now,  the  formula  r,     /~        1 

does  not  mean  that  the  independent  variable  must  be  denoted 
by  the  letter  x.  If  the  independent  variable  is  y,  the  formula 
reads : 

D,Vy  =  -i-- 

2Vy 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS     37 

Consequently  (1)  can  be  written  in  the  form : 
(2)  Z>xtt  =  -^(-2a>)  = 


2\A/  Va2  —  x1 

We  have,  then,  as  the  final  result : 


I),  Vet2  —  x1 


Va-  —  x2 


Example  2.     To  differentiate  the  function 

1 

y  = 

(1  -  Z)3 

Let  z  =  1  —  #. 

Then  y  =  z-3. 

To  apply  Theorem  V  in  the  present  case,  the  letters  u  and  y 
must  be  replaced  respectively  by  y  and  z.  Thus  Theorem  V 
reads  here  :  Dy  =  D^Djgf 

or  Dxy  =  Djr3Dx(l  -  x). 

Since  Formula  (8)  of  §  2  has  been  extended  to  negative  in- 
tegral values  of  n  by  §  7,  Ex.  2,  we  have : 

Z^z-3=-3z-4. 

Hence  Dxy  =  -  3  «-<  ( -  1)  =  - . 


or  Z> 


2!" 

1  3 


(i  -  xy    (i  -  xy 


EXERCISES 

Differentiate  the  following  functions  : 


1.  y=Va-+x\  Ans. 


Va2  +  x2 


1  A  X 

2.*  ?/= —  -4ws.    —  ■ 


Va2  —  x"-  V(a2  —  a2)3 

*  Note  that  Formula  (8)  of  §  2  has  also  been  shown  to  hold  for  the 
case  n  =  —  \  ;  §  4,  Ex.  1 . 


38  CALCULUS 


3.  y  =  Vl  +  x  +  x1.  Ans, 


5. 


?/  =  —  Ans. 

y/3-2x  +  4x2 


(1  -  x)3 


l  +  2x 

2  VI  +  x  +  x2 

l-2x 

V3-2x  +  4a;; 

J»    ^+1'- 

(2 -3  a;)2  (2 -3  a;)3 


a;2  0  (1-z)3 

7.    V  =  —         — •  8.    y  =  ± *-> 

(1+2  a;)4  H      (2  +  xf 


9.    y=[—Z—     ■  10.     u  = 


Kl  —  x)  1  —  2  a;  +  a;2 

11.*   it  =  x*(l  -  a:)4.  4ns.    (1  —  5a;)(l  —  a?)3. 

12.  u  =  x(a  +  bx)n.  Ans.    [a  +(n  +  l)&a;](a  +  bx)n~\ 

13.  m  =  a;2(a  +  &»)".  14.    u  =  x3(l  —  x)i. 

2a-3x 


15.        M  =  xVa  —  :f.  ^l/(.s. 


2  V«  —  x 


16.    ?( =  x2V«2  -  a;'2.  17.    u  =  x Vl  +  x  +  x2. 

18.    m  = ^ 19.    it  = 


Va2  —  a;2  Vl  +  x  +  .t- 

/« +  &a?  „,  .  1 

20.    u  =\/ — ! 21. t    w  =  —  — • 

Vc  +  dx-  (V      'Jujl')* 

22.    «  =  ————•  23.    w  = 


(a;'2  —  l)2  1  +  x  +  a;2 

a?-3&a£+3&2a;-&s         oc  a+b 

24.    W  =  —  25.    M=- 


6  —  x  (a  4-  6aj)2 

*  Use  Theorem  III. 

t  Do  not  use  Theorem  IV. 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS    39 

9.   Differentiation  of  Implicit  Algebraic  Functions.    When 
x  and  y  are  connected  by  such  a  relation  as 

x2  +  y2  —  a?, 

or  x3  —  2xy  +  y5  =  0, 

or  xy  sin  y  =  x  -\-y  log  x, 

i.e.  if  y  is  given  as  a  function  of  x  by  an  equation, 

P(x,  y)=0         or         4> (x,  y)=* (x,  y), 

which  must  first  be  solved  for  y,  then  y  is  said  to  be  an  implicit 
function  of  x.  If  we  solve  the  equation  for  y,  thus  obtaining 
the  equation 

y  =/(»), 

y  thereby  becomes  an  explicit  function  of  x. 

By  an  algebraic  function  of  x  is  meant  a  function  y  which 
satisfies  an  equation  of  the  form 

G(x,y)  =  0, 

where  G  (x,  y)  is  an  irreducible  polynomial  in  x  and  y ;  i.e.  a 
polynomial  that  cannot  be  factored  and  written  as  the  product 
of  two  polynomials. 

Thus  the  polynomials  are  algebraic  functions  ;  for  if 

y  =  a0  +  a^x  +  •  ■  •  +  anxn  =  P(x), 

then  y  satis  ties  the  algebraic  equation 

G(x,  y)=y-P(x)=0. 

Similarly,  the  fractions  in  x  are  algebraic  functions  ;  for  if 

where  P(x)  and  Q(x)  are  polynomials  having  no  common 
factor,  then  y  satisfies  the  algebraic  equation 

G(x}y)=Q(x)y-P(x)=0. 


40  CALCULUS 

The  polynomials  and  the  fractions  are  also  called  rational 

functions.     Thus, 

J  ax  +  by 

x-  +  y- 

is  a  rational  function  of  the  two  independent  variables  x  and  y. 
Again,  all  roots  of  polynomials,  as 

y  =  Vl  -f  x  +  x3, 
or  such  functions  as 


y 


*  1  —  x 


are  algebraic,  as  is  seen  on  freeing  the  equation  from  radicals 
and  transposing.  The  converse,  however,  —  namely,  that  every 
algebraic  function  can  be  expressed  by  means  of  rational  func- 
tions and  radicals,  —  is  not  true. 

In  order  to  differentiate  an  algebraic,  function,  it  is  sufficient 
to  differentiate  the  equation  as  it  stands.     Thus  if 

(1)  x2  +  y2  =  a2, 
we  have 

(2)  Dxx"  +  DJ-  =  Dxa2. 

To  find  the  value  of  the  second  term,  apply  Theorem  V,  §  8. 
Thus  Dzf-  =  DyfDxy  =  2yDzy. 

This  last  factor,  Dxy,  is  precisely  the  derivative  we  wish  to 
find,  and  it  is  given  by  completing  the  differentiations  indi- 
cated in  (2) :  _         _    _  . 
W                      2.x  +  2yDJcy  =  0, 

and  solving  this  equation  for  Dxy : 
Dxy  = 

y 

The  final  result  is,  of  course,  the  same  as  if  we  had  solved 

equation  (1)  foi»  y  : 
4  w  y=±y/a?-x*i 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS    41 

and  then  differentiated : 


Dxy  = 

±       ~X      - 

X 

Va2  -  x2 

y 

In  the 

case, 

however, 

of  the  equation 

(3) 

X 

3  -  2xy  +  if  =  0, 

we  cannot  solve  for  y  and  obtain  an  explicit  function  expressed 
in  terms  of  radicals.  Nevertheless,  the  equation  defines  y  as 
a  perfectly  definite  function  of  x ;  for,  on  giving  to  x  any 
special  numerical  value,  as  x  =  2,  we  have  an  algebraic  func- 
tion for  y,  —  here,  ,  _ 

J'  y5-±y  +  8  =  0, 

and  the  roots  of  this  equation  can  be  computed  to  any  degree 
of  precision. 

To  find  the  derivative  of  this  function,  differentiate  equation 

(3)  as  it  stands  with  respect  to  x : 

(4)  Dxx3  -  2 Dx(xy)  +  Dxif  =  0. 

The  second  term  in  this  last  equation  can  be  evaluated  by 

Theorem  III  of  §  7  :    _  .     N 

Dx(xy)  =  xDxy  +  y, 

where  Dxy  denotes  the  derivative  we  wish  to  find. 

To  the  evaluation  of  the  third  term  in  (4)  Theorem  V  of  §  8 

aPpli6S:  Dxf=5y*Dj, 

HenC?  3  x2  -  2  xDxy  -  2  y  +  5  y*Dxy  =  0. 

Solving  this  equation  for  Dxy,  we  have  as  the  final  result : 

n         2v-3z2 
xy      5y*-2x 

Thus,  for  example,  the  curve  is  seen  to  go  through  the  point 
(1,  1),  and  its  slope  there  is 


42  CALCULUS 

The  differentiation  of  implicit  functions  as  set  forth  in  the 
above  examples  is  based  on  the  assumptions  a)  that  the  given 
equation  defines  ?/asa  function  of  x ;  b)  that  this  function  has 
a  derivative.  The  proof  of  these  assumptions  belongs  to  a 
more  advanced  stage  of  analysis.  In  the  case,  however,  of  the 
equations  we  meet  in  practice,  —  for  example,  such  equations 
as  come  from  a  problem  in  geometry  or  physics,  —  the  condi- 
tions for  the  existence  of  a  solution  and  of  its  derivative  are 
fulfilled,  and  we  shall  take  it  for  granted  henceforth  that  this 
is  true  of  the  implicit  functions  we  meet. 

Derivative  of  x'\  n  Fractional.     We  are  now  in  a  position  to 

prove  the  theorem  _  , 

JJxxn  =  nxn~l 

for  the  case  that  n  is  a  fraction.     Let 

Q 

where  p,  q  are  whole  numbers  which  are  prime  to  each  other. 

Let 

p 

y  =  x". 
Then  y  =  .r". 

Differentiating  each  side  of  this  equation  with  respect  to  x, 

we  have : 

Dzy  =  Dtx*, 

and  since,  by  Theorem  V,  §  8. 

Dxy>  =  Dyy«Dxy  =  qy'^D^, 

it  follows  that  _ 

qy-lDxy  =  p.r    '. 

or  Dji  —  ±-  —  • 

qy"~l 

This  last  denominator  has  the  value 

p  p-- 

(a«)»-1  =  cc    «. 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS     43 

Hence  * — -  =  ' —  =  x? 

irl      P-p- 

x    q 
We  see,  then,  that 

D,y  =  £■  x"1  —  naf-K  q.  e.  d. 

If,  finally,  n  is  a  negative  fraction,  n  =  —  ??i,  the  proof  can 
be  given  precisely  as  was  done  in  §  7,  Ex.  2.     Thus  the  theorem 

Djxn  —  nxn~1 

is  now  established  for  all  commensurable  values  of  n. 

The  theorem  is  true  even  when  n  is  irrational,  e.g.  n  =  n  or 
V2;  the  proof  depends  on  the  logarithmic  function  and  will 
be  given  when  that  function  has  been  differentiated. 

Example.     Differentiate  the  function 


Apply  Theorem  V,  §  8,  setting 

z  =  a?  —  x3. 

Then  Dxy  =  D  cz*  =  Dzz*Dxz  =  |  z_*(  -  3  x*) . 


Hence  Dx  vc 


x~ 


V  (a3  —  x3)2 


EXERCISES 


2  x3  —  3  xhj  +  4  xy  +  6  y3  =  0, 
?/4  -  2  z?/2  =  a,*4, 


3.    Show  that  the  curve 

xi  —  2  xy1  +  y3  +  3  x  —  3  ?/  =  0 
cuts  the  axis  of  x  at  the  origin  at  an  angle  of  45°. 


1. 

If 

find 

Dxy 

2. 

If 

find  Dxy 

44  CALCULUS 

4.   Plot  the  curve       x*  4-  y*  =  81, 

taking  1  cm.  as  the  unit.  Show  that  this  curve  is  cut  orthog- 
onally by  the  bisectors  of  the  angles  made  by  the  coordinate 
axes. 

Differentiate  the  following  functions  : 

-1 


5.    u  =  Vl  —  x.  Ans. 


6.    «=Va2-2  ax  +  x2.  A  n  s 


5V(l-a)4 
-2 


7.    11  =  Vc3  —  3c2x  4-  Sex'1  —  .j-3.      Ans. 


3V  a  —  x 
3 


5  Vc2  —  2cx  +  x2 


I       &  4  ' " 

8.    u  =  \ 9.    u  =  x\  a  4-  bx  4-  ex2. 

M  1  —a 

10.    m=- — Ans. 


3Vz2(i  -  xY 


,.      Va  —  a;  4-  Va  4-  a;  .         a2  4-  aVa2  —  &2 

Va  —  k  —  Va  +  x  x-  y/a2  —  x2 

12.   y  =  v/ax2.  13.    r=V«0. 

1  —  aT*                                          ,         n         7-2Val 
14.    u= -•  ^l»s.    D.u  = 

X*  lOv'S" 


1+_0J2 

s/x 


15.   y  =     "J*     ■  16.    h=#V2x. 


17.  o  =  4>4- 1- 

V* 

18.  (y2  +  i)V^^.  Ans.    7y4~2y2~1. 

2  \  ,v3  -  y 


19.  (*2-«2)f,  ^s.  3ft2vV~a2. 

S3  $< 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS     45 

20.  a~X       ■  21.     4±~X   ■ 


22.    y  =  x(ar  —  x°-y.  23.    u=(b  —  t)y/b  +  t. 

24.  Find  the  slope  of  the  curve  y  =  x*  in  the  point  whose 
abscissa  is  2.  Ans.    tan  t  =  .115. 

25.  IipvlA=c,      find      Dt^. 

26.  If  y-\/x  =  1  +  .r,       find      Dxy.  Ans.    — 


2  xs/x 

27.  Differentiate  y  in  two  ways,  where 

xy  +  4  y  =  3  a;, 
and  show  that  the  results  agree. 

28.  The  same,  when  y1  =  2  mx. 

29.  Show  that  the  curves 

3y  =  2 x  4-  ary ,  2 y  +  3  a;  4-  //5  =  a;3?/, 

intersect  at  right  angles  at  the  origin. 

30.  Find  the  angle  at  which  the  curves 

2x  =  xi  —  xy  4-  x5,  x4  +  ?/4  4-  5a;  =  ly, 

intersect  at  the  origin.  Ans.   tan  <f>  =  1.4. 


CHAPTER    ill 


APPLICATIONS 

1.   Tangents  and  Normals.     By  the  tangent  line,  or  simply 
the  tangent,  to  a  curve  at  any  one  of  its  points,  P,  is  meant  the 

straight  line  through  P, 
whose  slope  is  the  same  as 
that  of  the  curve  at  that 
point. 

Let  the  coordinates  of  P 
be  denoted  by  (x0,  y0).    Now, 
the  equation  of  the  straight 
line  through  P,  whose  slope  is  A,  is 

.'/  —  2/o  =  Kx  -  xo)- 

On  the  other  hand,  the  slope  of  the  curve  at  any  point  is  Dxy. 
If  we  denote  the  value  of  this  slope  at  (x0,  y0)  by  (Dj/)0,  this 
will  be  the  desired  value  of  \ : 

Hence  the  equation  of  the  tangent  to  the  curve 

y=f(x)         or         F(x,y)=0 
at  the  point  (.r0,  y0)  is 
(1)  y-!/o=(DIy)o(x-x0). 

Since  the  normal  is  perpendicular  to  the  tangent,  its  slope, 
A',  is  the  negative  reciprocal  of  the  slope  of  that  line,  or 

1 


X'=- 


46 


APPLICATIONS  47 

Hence  the  equation  of  the  normal  to  the  curve  at  (x0,  y0)  is 
(2)     y-yQSSf-——(x-a^     or     x-x0  +  (Dzy)0  .  (y  -  y0)  =  0. 


Example  1.     To  find   the  equation  of  the  tangent   to  the 

curve 

y  ^x3 

in  the  point  x  =  \,  y1^^.     Here 

Dxy  =  3x*;        (Ay)o  =  [3aj2]^  =  f. 
Hence  the  equation  of  the  tangent  is 

y-i  =  l(*-i)  or  3*-4y-l  =  0. 

Example  2.     Let  the  curve  be  an  ellipse : 

a2     ft2 
Differentiating  the  equation  as  it  stands,  we  get : 

2f+ffZ^  =  0,  Dxy=-h^- 

a2      o-  a2y 

Hence  the  equation  of  the  tangent  is 

0  Xn  /  \ 

y-yo  =  -^(x-x0). 

This  can  be  transformed  as  follows : 

a2yoy  —  «2#o2  =  —  Vx0x  -+-  b2x02, 
b2x0x  +  a2y0y  =  a2y02  +  b2x02  =  a2b\ 

a2       b2 


2 

circle 


48  CALCULUS 

EXERCISES 

1.   Find  the  equation  of  the  tangent  of  the  curve 

y  =  x3  —>x 

at  the  origin ;  at  the  point  where  it  crosses  the  positive  axis 
of  x.  Ans.    x  +  y  =  0 ;  2x  —  y  —  2  =  0. 

.    Find  the  equation  of  the  tangent  and  the  normal  of  the 

Q 

a2  +  2/2  =  4 
at  the  point  (1,  V3)  and  check  your  answer. 

3.  Show  that  the  equation  of  the  tangent  to  the  hyperbola 

a2     b2 
at  the  point  (x0,  ya)  is 

a2       b2 

4.  Show  that  the  equation  of  the  tangent  of  the  parabola 

?/2  =  2mx 
at  the  point  (x0,  y0)  is 

yQy  =  m(x  +  x0). 

5.  Show  that  the  equation  of  the  tangent  of  the  parabola 

y2  =  m2  —  2  mx 
at  the  point  (x0,  y0)  is 

y0y  =  m2  —  m(x+x0). 

6.  Show  that  the  equation  of  the  tangent  of  the  equilateral 

hyperbola 

xy  =  a2 
at  the  point  (x0,  y0)  is 

ytix  +  x0y  =  2a2. 

7.  Find  the  equation  of  the  tangent  to  the  curve 

x3  +  y3  =  a\x  —  y) 
at  the  origin.  Ans.   x  =  y. 


APPLICATIONS 


49 


8.  Show  that  the  area  of  the  triangle  formed  by  the  coordi- 
nate axes  and  the  tangent  of  the  hyperbola 

xy  =  a2 
at  any  point  is  constant. 

9.  Find  the  equation  of  the  tangent  and  the  normal  of  the 

curve 

xb  =  a3y2 

in  the  point  distinct  from  the  origin  in  which  it  is  cut  by  the 
bisector  of  the  positive  coordinate  axes. 

10.  Show  that  the  portion  of  the  tangent  of  the  curve 

Xs  +  y*  =  a3" 

at  any   point,   intercepted   between   the   coordinate   axes,   is 
constant. 

11.  The  parabola  y2  =  2  ax  cuts  the  curve 

x3  —  3  axy  +  ?/  =  0 

at  the  origin  and  at  one  other  point.     Write  down  the  equa- 
tion of  the  tangent  of  each  curve  in  the  latter  point. 

12.  Show  that  the  curves  of  the  preceding  question  intersect 
in  the  second  point  at  an  angle  of  32°  12'. 

2.  Maxima  and  Minima.  Problem.  From  a  piece  of  tin 
3  ft.  square  a  box  is  to  be  made  by  cutting  out  equal  squares 
from  the  four  corners  and 
bending  up  the  sides.  Deter- 
mine the  dimensions  of  the 
box  of  this  description  which 
will  hold  the  most.  3— 2x 




1 

J 

Z-2x 


Fig.  10 


Solution.     Let    x    be     the 
length    of    the    side    of    the 
square     removed;     then    the 
dimensions  of  the  box  are  as  indicated  in  the  diagrams.     De- 
noting the  cubical  content  of  the  box  by  u,  we  have  : 


50 


CALCULUS 


1) 


u  =  x(3-2xf, 


2) 


u  =  9x-  12x2  +  4a?>. 


The  problem  is,  then,  to  find  the  value  of  x  which  makes  u 
as  large  as  possible,  x  being  restricted  from  the  nature  of  the 
case  to  being  positive  and  less  than  f : 


3) 


0  <  x  < 


The   problem  can  be  treated   graphically  by  plotting  the 
curve  1).     We  wish  to  find  the  highest  point  on  this  curve. 

It  appears  to  be  the  point 
for  which  x  =  \,  u  =  2, 
since  other  values  of  x 
which  have  been  tried  lead 
to  smaller  values  of  u. 

The  foregoing  method 
has  the  advantage  that  it 
is  direct,  for  it  assumes  no 
knowledge  of  mathematics 
beyond  curve  plotting.  It 
lias  the  disadvantage  that 
curve  plotting,  even  in  the 
simplest  cases,  is  labori- 
ous ;  and,  furthermore,  we 
have  not  really  proved 
that  x  =  .',  is  the  best 
value.     We  have  merely  failed  to  find  a  better  one. 

The  Calculus  supplies  a  means  of  meeting  both  the  difficul- 
ties mentioned,  and  yielding  a  solution  with  the  greatest  ease. 
The  problem  is  to  find  the  highest  point  on  the  curve.  At. 
this  point,  the  tangent  of  the  curve  is  evidently  parallel  to 
the  axis  of  x.  Consequently,  the  slope  of  the  tangent,  i.e. 
tan  t  =  Dxu,  must  have  the  value  0  here  : 

Dm  =  0. 


Fig 


APPLICATIONS  51 

All   we   need   do,   therefore,   is   to   compute   Dxu,  most  con- 
veniently from  equation  2),  and  set  the  result  equal  to  0 : 

Da%  =  9-24aj  +  12aj2  =  0. 

On  solving  this  quadratic  equation  for  x,  we  find  two  roots, 


Only  one  of  these,  however,  lies  within  the  range  3)  of  possible 
values  for  x,  namely,  the  value  x  =  i,  and  hence  this  is  the 
required  value. 

EXERCISES 

1.  Work  the  foregoing  problem  for  the  case  that  the  tin  is 
a  rectangle  1  by  2  ft. 

Plot  accurately  the  graph,  taking  10  cm.  as  the  unit,  and 
determine  in  this  way  what  appears  to  be  the  best  value  for  x, 
correct  to  one  eighth  of  an  inch. 

Solve  the  problem  by  the  Calculus,  and  show  that  the  best 
value  for  x  is  .21132  ft.,  or  2.5359  in. 

2.  A  farmer  wishes  to  fence  off  a  rectangular  pasture  along 
a  straight  river/  one  side  of  the  pasture  being  formed  by  the 
river  and  requiring  no  fence.  He  has  barbed  wire  enough  to 
build  a  fence  1000  ft.  long.  What  is  the  area  of  the  largest 
pasture  of  the  above  description  which  he  can  fence  off  ? 

3.  Show  that,  of  all  rectangles  having  a  given  perimeter, 
the  square  has  the  largest  area. 

4.  Show  that,  of  all  rectangles  having  a  given  area,  the 
square  has  the  least  perimeter. 

5.  Each  side  of  a  shelter  tent  is  a  rectangle 
G  x  8  f t.     How  must  the  tent  be  pitched  so  as  to 
afford  the  largest  amount  of  room  inside  ?     The         ^^  ^ 
ends  are  to  be  open. 

Ans.  The  angle  along  the.  ridge-pole  must  be  a  right 
angle. 


52  CALCULUS 

6.  Divide  the  number  12  into  two  parts  such  that  the  sum 
of  their  squares  may  be  as  small  as  possible. 

(What  is  meant  is  such  a  division  as  this  :    one  part  might 

be  4,  and  then  the  other  would  be  8.     The  sum  of  the  squares 

would  here  be  16  +  64  =  80.) 

* 

7.  Divide  the  number  8  into  two  such  parts  that  the  sum 
of  the  cube  of  one  part  and  twice  the  cube  of  the  other  may  be 
as  small  as  possible. 

8.  Divide  the  number  9  into  two  such  parts  that  the 
product  of  one  part  by  the  square  of  the  other  may  be  as  large 
as  possible. 

9.  Divide  the  number  8  into  two  such  parts  that  the  product 
of  one  part  by  the  cube  of  the  other  may  be  as  large  as  possible. 

10.  At  noon,  one  ship,  which  is  steaming  east  at  the  rate 
of  20  miles  an  hour,  is  due  south  of  a  second  ship  steaming 
south  at  16  miles  an  hour,  the  distance  between  them  being 
82  miles.  If  both  ships  hold  their  courses,  show  that  they 
will  be  nearest  to  each  other  at  2  p.m. 

11.  If,  in  the  preceding  problem,  the  second  ship  lies  to 
from  noon  till  one  o'clock,  and  then  proceeds  on  her  southerly 
course  at  16  miles  an  hour,  when  will  the  ships  be  nearest  to 
each  other? 

12.  Find  the  least  value  of  the  function 

y  =  x2  +  6  x  +  10.  Ans.    1. 

13.  What  is  the  greatest  value  of  the  function 

y  =  3  x  —  x3 
for  positive  values  of  x  ? 

14.  For  what  value  of  x  does  the  function 

12VJ; 

l  +  4x 

attain  its  greatest  value '.'  Ans.   x=\. 


APPLICATIONS 


53 


15.    At  what  point  of  the  interval  a  <  x  <  b,  a  being  posi- 
tive, does  the  function 


x 


(x  -~  a)(b  —  x) 
attain  i#s  least  value  ?  Ans.   x  =  Vo&. 

16.  Find  the  most  advantageous  length  for  a  lever,  by 
means  of  which  to  raise  a 
weight  of  490  lb.  (see  Fig.  13), 
if  the  distance  of  the  weight 
from  the  fulcrum  is  1  ft.  and  the 
lever  weighs  5  lb.  to  the  foot. 

3.  Continuation:  Auxiliary  Variables.  It  frequently,  —  in 
fact,  usually,  —  happens  that  it  is  more  convenient  to  formu- 
late a  problem  if  more  variables  are  introduced  at  the  outset 
than  are  ultimately  needed.  The  following  examples  will 
serve  to  illustrate  the  method. 

Example  1.  Let  it  be  required  to  find  the  rectangle  of 
greatest  area  which  can  be  inscribed  in  a  given  circle. 

It  is  evident  that  the  area  of  the 
rectangle  will  be  small  when  its  alti- 
tude is  small  and  also  when  its  base 
is  short.  Hence  the  area  will  be 
largest  for  some  intermediate  shape. 
Let  u  denote  the  area  of  the  rec- 
tangle.    Then 

(1)  u  =  4  xy. 

But  x  and  y  cannot  both  be  chosen 
arbitrarily,  for  then  the  rectangle 
will  not  in  general  be  inscriptible  in  the  given  circle.  In  fact, 
it  is  clear  from  the  Pythagorean  Theorem  that  x  and  y  must 
satisfy  the  relation : 

(2)  a2  +  y2  =  a2. 

We  could  now  eliminate  y  between  equations  (1)  and  (2), 
thus  obtaining  u  in  terms  of  x  alone ;  and  it  is,  indeed,  im- 


Fig.  14 


54  CALCULUS 

portant  to  think  of  this  elimination  as  performed,  for  there  is 

only  one  independent  variable  in  the  problem.     The  graph  of 

u,  regarded  as  a  function  of  x,  starts  at  the  origin,  rises  as  x 

increases,  but  finally  comes  back  to  the  axis  of  x  ag^n  when 

x  =  a.     All   this   we  read  off,  either  from  the 

meaning  of  u  and  x  in  the  problem  or  from 

Jf   equations  (1)  and  (2). 

Tt  is  better,  however,  in  practice  not  to  eliiui- 
Fig.  15  . 

nate  y,  but  to  differentiate  equations  (1)  and 

(2)  with  respect  to  x  as  they  stand,  and  then  set  Dxu  =  0. 

Thus  from  (1), 

Dxu  =  4(y+xDxy)=Q, 

and  from  (2),  2x  +  2  yDzy  =  0. 

From  the  second  of  these  equations  we  see  that 

y 

Substituting  for  Dzy  this  value  in  the  first  equation,  we  get : 

x2 
y =  0         or         y2  =  x2. 

y 

Since  x  and  y  are  both  positive  numbers,  it  follows  that 

y  =  «■ 
Hence  the  maximum  rectangle  is  a  square. 

EXERCISES 

1.  Work   the  same   problem   for  an  ellipse,  instead   of  a 
circle. 

2.  Work  the  problem  for  the  case  of  a  variable  rectangle 
inscribed  in  a  fixed  equilateral  triangle. 

Example  2.     To  find  the  most  economical  dimensions  for  a 
tin  dipper,  to  hold  a  pint. 


APPLICATIONS  55 

Here,  the  amount  of  tin  required  is  to  be  as  small  as  pos- 
sible, tbe  content  of  the  dipper  being  given.  Let  u  denote  the 
surface,  measured  in  square  inches.     Then 

a)  u  =  2irrh  +  ttt-. 

But  r  and  h  cannot  both  be  chosen  arbitrarily,  for 

then  the  dipper  would  not  in  general  hold  a  pint. 

If  V  denotes  the  given  volume,  measured  in  cubic        Fig.  16 

inches,  then,  since  this  volume  can  also  be  expressed  as  7rr2h, 

we  have 

6)  7rr%  =  V. 

Differentiating  equation  a)  with  respect  to  r  and  setting 
Dru  =  0,  we  have  : 

DTu  =  7T  \2h  +  2rDJi  +  2rl  =  0, 
or  J         ' 

c)  h  +  rDrh  +  r  =  0. 
Differentiating  b)  we  get : 

d)  7r\2rh  +  rWrh\  =  0. 

Now,  r  cannot  =  0  in  this  problem,  and  so  we  may  divide  this 
last  equation  through  by  r,  as  well  as  by  it  : 

e)  2h  +  rDTh  =  0. 

It  remains  to  eliminate  Drh  between  equations  c)  and  e). 
From  e), 

2  h 
Drh  =  -—. 
r 

Substituting  this  value  of  DTh  in  c),  we  find : 

/)  h-2h  +  r  =  0,         or        r  =  h. 

Hence  the  depth  of  the  dipper  must  just  equal  its  radius. 

Discussion.  Just  what  have  we  done  here  ?  The  steps  we 
have  taken  are  suggested  clearly  enough  by  the  solution  of 


56  CALCULUS 

Example  1.  We  have  chosen  one  of  the  two  variables,  r  and 
h,  as  the  independent  variable  (here,  r) ;  differentiated  the 
function  u,  which  is  to  be  made  a  minimum,  with  respect  to  r, 
and  set  Dru  =  0.  Then  we  differentiated  the  second  equation 
6),  likewise  with  respect  to  r,  eliminated  Dru,  and  solved. 
But  what  does  it  all  mean  ?     What  is  behind  it  all  ? 

Just  this :  the  quantity  u,  in  the  nature  of  the  case,  is  a 
function  of  r.  For,  when  to  r  is  given  any  positive  value,  a 
dipper  can  be  constructed  which  will  fulfill  the  requirements. 
Now,  if  r  is  very  large,  we  shall  have  a  shallow  pan,  and  evi- 
dently the  amount  of  tin  required  to  make  it  will  be  large ;  — 
i.e.  u  will  also  have  a  large  value. 

But  what  if  r  is  small  ?  We  shall  then  have  a  high  cylinder 
of  minute  cross  sections,  i.e.  a  pipe.  Is  it  clear  that  a,  the 
surface,  will  be  large  in  this  case,  too?  I  fear  not,  for  it  is 
purely  a  relative  question  as  to  how  high  such  a  pipe  must  be 
to  hold  a  pint,  and  I  see  no  way  of  guessing  intelligently. 
By  means  of  equation  b),  however,  we  see  that 

77- r- 

and  if  we  substitute  this  value  in  a),  we  get 

u\  I  V  IV 

u  =  2irr h  nr2  = (-  ir)'2. 

irr"1  r 

From  this  last  formula  it  is  clear  that, 
when  r  is  small,  u  actually  is  large. 

The  graph  of  u,  regarded  as  a  func- 
tion of  r,  is  therefore  in  character  as 
shown   by   the   accompanying   figure. 

„  It  is  a  continuous  curve  lying  above 

Fig.  17  J      & 

the  axis  of  r,  very  high  when  r  is  small, 

and   also   very  high  when  r  is   large.     It  has,   therefore,   a 

lowest  point,  and  for  this  value  of  r,  the  area  u  of  the  dipper 

will  be  least.     But  at  this  lowest  point  the  slope  of  the  curve, 

Dru,  has  the  value  0.     Thus  we  see,  first,  that  we  have  a  genu- 


APPLICATIONS  57 

ine  minimum  problem ;  —  there  is  actually  a  dipper  of  small- 
est area.  Secondly,  equations  c)  and  d)  must  hold,  and  since 
from  these  equations  it  follows  by  elimination  that  r  =  h,  there 
is  only  one  such  dipper,  and  its  radius  is  equal  to  its  altitude. 
The  problem  is,  then,  completely  solved. 

We  inquired  merely  for  the  shape  of  the  dipper.  If  the  size 
had  been  asked  for,  too,  it  could  be  found  by  solving  b)  and  f) 
for  r  and  h,  and  expressing  V  in  cubic  inches  : 

v=  2|i  =  28.875, 

7rr3  =  28.87,  r  =  2.095. 

It  can  happen  in  practice  that  a  function  attains  its  greatest 
or  its  least  value  at  the  end  of  the  interval.  In  that  case,  the 
derivative  does  not  have  to  vanish.  Usu- 
ally, the  facts  are  patent,  and  so  no  special 
investigation  is  needed.  But  it  is  neces- 
sary to  assure  oneself  that  a  given  problem 
which  looks  like  one  of  the  above  does  not 


a 
Fig.  18 


come  under  this  head,  and  this  is  done,  as 
in  the  cases  discussed  in  the  text,  by  show- 
ing that  near  the  ends  of  the  interval  the  values  of  the  func- 
tion are  larger,  for  a  minimum  problem,  than  for  values  well 
within  the  interval. 

EXERCISE 

Discuss  in  a  similar  manner  the  best  shape  for  a  tomato  can 
which  is  to  hold  a  quart.  Here,  the  tin  for  the  top  must  also 
be  figured  in.  Show  that  the  height  of  such  a  can  should  be 
equal  to  the  diameter  of  the  base.  As  to  the  size  of  the  can, 
its  height  should  be  4.19  in. 

A  General  Remark.  It  might  seem  as  if  the  method  used  in 
the  solution  of  the  above  problems  were  likely  to  be  insecure, 
since  the  graph  of  such  a  function  u  might,  in  the  very  next 
problem,  look  like  the  accompanying  figure.  In  such  a  case, 
there  would  be  several  values  of  x,  for  each  of  which  Dxii  =  0, 


58 


CALCULUS 


and  we  should  not  know  which  one  to  take.  Curiously  enough, 
this  case  does  not  arise  in  practice,  —  at  least,  I  have  never 
come  across  a  physical  problem  which 
led  to  this  difficulty.  In  problems 
like  the  above,  there  must  be  at  least 
one  x  for  which  D.n  =  0 ;  and  when 
we  solve  a  given  problem,  we  actually 
q  find  only  one  x  which  fulfills  the  con- 

Fiq.  19  dition.     Thus  there  is  no  ambiguity. 


EXERCISES 

1.  A  300-gallon  tank  is  to  be  built  with  a  square  base  and 
vertical  sides,  and  is  to  be  lined  with  copper.  Find  the  most 
economical  proportions. 

Ans.  The  length  and  breadth  must  each  be  double  the 
height. 

2.  Find  the  cylinder  of  greatest  volume  which 
can  be  inscribed  in  a  given  cone  of  revolution. 

Ans.    Its  altitude  is  one-third  that  of  the  cone. 

3.  What  is  the  cylinder  of  greatest  convex 
surface  that  can  be  inscribed  in  the  same  cone  ? 

Ans.    Its  altitude  is  half  that  of  the  cone. 

4.  Of  all  the  cones  which  can  be  inscribed  in  a  given  sphere, 
find  the  one  whose  lateral  area  is  greatest. 

Ans.  Its  altitude  exceeds  the  radius  of  the  sphere  by  33^  % 
of  that  radius. 

5.  Find  the  volume  of  the  greatest  cone  of  revolution  which 
can  be  inscribed  in  a  given  sphere. 

6.  If  the  top  and  bottom  of  the  tomato  can  considered  in 
the  Exercise  of  the  text  are  cut  from  sheets  of  tin  so  that  a 
regular  hexagon  is  used  up  each  time,  the  waste  being  a  total 
loss,  what  will  then  be  the  most  economical  proportions  for 
the  can  ? 


APPLICATIONS  59 

7.  If  the  strength  of  a  beam  is  proportional  to  its  breadth 
and  to  the  square  of  its  depth,  find  the  shape  of  the  strongest 
beam  that  can  be  cut  from  a  circular  log. 

Ans.   The  ratio  of  depth  to  breadth  is  V2. 

8.  Assuming  that  the  stiffness  of  a  beam  is  proportional  to 
its  breadth  and  to  the  cube  of  its  depth,  find  the  dimensions 
of  the  stiffest  beam  that  can  be  sawed  from  a  log  one  foot  in 
diameter. 

9.  What  is  the  shortest  distance  from  the  point  (10,  0)  to 
the  parabola  2  _  ax9 

10.  What  points  of  the  curve 

y2  =  x3 
are  nearest  (4,  0)  ? 

11.  A  trough  is  to  be  made  of  a  long  rectangular-shaped 
piece  of  copper  by  bending  up  the  edges  so  as  to  give  a  rec- 
tangular cross-section.  How  deeu  should  it  be  made,  in  order 
that  its  carrying  capacity  may  be  as  great  as  possible  ? 

12.  Assuming  the  density  of  water  to  be  given  from  0°  to 
30°  C.  by  the  formula 

p  =  Po(l  +  at  +  pt2  +  yf), 

where  p0  denotes  the  density  at  freezing,  t  the  temperature, 
and 

a  =5.30  x  10-5,         p  =  -  6.53  x  10"6,         y  =  1.4  X  10"8, 

show  that  the  maximum  density  occurs  at  t  =  4.08°. 

13.  Tangents  are  drawn  to  the  arc  of  the  ellipse 

&,£  =  1 

a2      52 

which  lies  in  the  first  quadrant.     Which  one  of  them  cuts  off 
from  that  quadrant  the  triangle  of  smallest  area  ? 

14.  Work  the  same  problem  for  the  parabola 

y2  =  a2  —  4  ax. 


60 


CALCULUS 


15.  Show  that,  of  all  circular  sectors  having  the  same  perim- 
eter, that  one  has  the  largest  area  for  which  the  sum  of  the 
two  straight  sides  is  equal  to  the  curved  side. 

4.  Increasing  and  Decreasing  Functions.  The  Calculus 
affords  a  simple  means  of  determining  whether  a  function  is  in- 
creasing or  decreasing  as  the  independent  variable  increases. 

Since  the  slope  of  the 
graph  is  given  by  Dxy, 
we  see  that  when  Dxy 
is  positive,  y  increases 
as  x  increases,  but 
when  Dxy  is  negative, 
y  decreases  as  x  in- 
Figure  21  shows  the  graph  in  general  when  Djf  is 


y 


0 


o 

Fig.  21 


creases, 
positive. 

In  each  figure  both  x  and  y  have  been  taken  as  positive. 
But  what  is  said  above  in  the  text  is  equally  true  when  one  or 
both  of  these  variables  are  negative ;  for  the  words  increase 
and  decrease  as  here  used  mean  algebraic,  not  numerical,  in- 
crease or  decrease.  Thus  if  the  temperature  is  ten  degrees  below 
zero  (i.e.  — 10°)  and  it  changes  to  eight  below  (—  8°),  we  say 
the  temperature  has  risen.  If 
we  measure  the  time  t,  in  hours 
from  noon,  then  10  a.m.  will 
correspond  to  t  =  —  2.  Let  u 
denote  the  temperature,  meas- 
ured in  degrees.  Then  a  tem- 
perature chart  for  24  hours 
from  midnight  to  midnight  might  look  like  the  accompanying 
figure.  At  any  instant,  t  =  t',  for  which  the  slope  of  the  curve, 
Dtu,  is  positive,  the  temperature  is  rising,  no  matter  whether 
the  thermometer  is  above  zero  or  below,  and  no  matter  whether 
t  is  positive  or  negative  ;  and  similarly,  when  Dtu  is  negative, 
the  temperature  is  falling. 

Again,  suppose  the  amount  of  business  a  department  store 


Fig.  22 


APPLICATIONS  61 

does  in  a  year,  as  represented  by  the  net  receipts  each  day,  be 
plotted  as  a  curve  (y  =  receipts,  measured  in  dollars ;  x  = 
time,  measured  in  days),  the  curve  being  smoothed  in  the 
usual  way.  Then  a  point  of  the  curve  at  which  the  derivative 
is  positive  (i.e.  Dzy  >  0)  indicates  that,  at  that  time,  the  busi- 
ness of  the  firm  was  increasing ;  whereas  a  point  at  which 
Dxy  <  0  means  that  the  business  was  falling  off. 

We  can  state  the  result  in  the  form  of  a  general  theorem, 
the  proof  of  which  is  given  by  inspection  of  the  figure  (Fig.  21) 
and  the  other  forms  of  the  figure,  brought  out  in  the  above 
discussion. 

Theorem  :    When  x  increases,  then 

(a)  if  Dxy  >  0,  y  increases; 

(b)  if  Dxy  <  0,  y  decreases.  < 

Application.  As  an  application  consider  the  condition  that 
a  curve  y  =f(x)  have  its  concave  side  turned  upward,  as  in 
Fig.  23.  The  slope  of  the  curve  is 
a  function  of  x  : 

tan  r  =  4>(x)- 


For,  when  x  is  given,  a  point  of 
the  curve,  and  hence  also  the  ^ 
slope  of  the  curve  at  this  point,  is 
determined.  Consider  the  tangent  line  at  a  variable  point  P. 
If  we  think  of  P  as  tracing  out  the  curve  and  carrying  the 
tangent  along  with  it,  the  tangent  will  turn  in  the  counter 
clock-wise  sense,  the  slope  thus  increasing  algebraically  as  x 
increases,  whenever  the  curve  is  concave  upward.  And  con- 
versely, if  the  slope  increases  as  x  increases,  the  tangent  will 
turn  in  the  counter  clock-wise  sense  and  the  curve  will  be  con- 
cave upward.     Now  by  the  above  theorem,  when 

Dx  tan  t  >  0, 

tanT  increases  as  x  increases.     Hence  the  curve   is   concave 
upward,  when  Dx  tan  t  is  positive ;  and  conversely. 


Fig.  23 


62 


CALCULUS 


The  derivative  Dx  tan  t  is  the  derivative  of  the  derivative 
of  y.  This  is  called  the  second  derivative  of  y,  and  is  denoted 
as  follows:  n/n   >     ,n, 

(read  :  "  D  x  second  of  y  ").* 

The  test  for  the  curve's  being  concave  downward  is  obtained 
in  a  similar  manner,  and  thus  we  are  led  to  the  following 
important  theorem. 

Test  fob  a  Curve's  being  Concave   Upward,  etc.     Hie 

curve  .,  N 

is  concave  upward  ivhen     Dx-y  >  0 ; 

concave  downward        ichen     Df-y  <  0. 

y  A  point  at  which 

the  curve  changes 
from  being  concave 
upward  and  be- 
comes concave 
downward  (or  vice 
versa)  is  called  a 
point  of  inflection. 
Since  Dx2y  changes 
sign  at  such  a  point, 

this   function   will   necessarily,    if   continuous,  vanish   there. 

Hence : 

A  necessary  condition  for  a  point  if  infection  is  that 
D*y  =  0. 

Example.     Consider  the  curve 

y  =  x3  —  3x. 

*  The  derivative  of  the  second  derivative,  Dx(Dx2y),  is  called  the  third 
derivative  and  is  written  Dx3y,  and  so  on. 


Fig.  24 


APPLICATIONS 


63 


Its  slope  at  any  point  is  given  by  the  equation 
Dzy  =  3  a;2  -  3. 

The  second  derivative  of  y  with 
respect  to  x  has  the  value 

D*y  =  6x. 

Thus  we  see  that  this  curve  is 
concave  upward  for  all  positive 
values  of  x,  and  concave  down- 
ward for  all  negative  values.  In 
character  it  is  as  shown  in  the  accompanying  figure. 


Fig.  25 


EXERCISES 

For  what  values  of  x  are  the  following  functions  increasing '.' 
For  what  values  decreasing  ? 

1.  ?/  =  4-2a,-2. 

2.  y  =  x*  —  2x  +  3. 

Ans.    Increasing,  when  x  >  1 ;  decreasing,  when  x  <  1. 

3.  y  =  5  +  12x-x*. 

4.  y  =  x3  —  27  x  +  7. 

Ans.    Increasing,  when  x  >  3,  and  when  x  <  —  3  ;    decreas- 
ing, when  —  3  <  x  <  3. 

5.  y  =  5  -f  6x  —  x3.  6.    y  =  x  —  xh. 

7.  y  =  x3  —  9x2  +  12x-l. 

In  what  intervals  are  the  following  curves  concave  upward ; 
in  what,  downward  ? 

8.  y  =  x3  —  3x2  -f-  7x  —  5. 

-4«s.    Concave   upward,   when   x  >  1 ;    concave   downward, 
when  x  <  1. 


1 


64  CALCULUS 

9.  y  =  15  +  8x  +  3x*-x3.         10.    y  =  z3  -  Qz2  -  x  -  1. 

11.  y  =  3  —  9x  +  2±x*-4x3.       12.   y  =  2xi-x\ 

13.  y  =  x*  —  4X3  -  6x  +  11.         14.    y  =  —121x  +  7a?  —  x7. 
15.  t/=13  +  23x-24a;2-r-l2a-3-z4. 

5.  Curve  Tracing.  In  the  early  work  of  plotting  curves 
from  their  equations  the  only  way  we  had  of  finding  out  what 
the  graph  of  a  function  looked  like  was  by  computing  a  large 
number  of  its  points.  We  are  now  in  possession  of  powerful 
methods  for  determining  the  character  of  the  graph  with 
scarcely  any  computation.  For,  first,  we  can  find  the  slope  of 
the  curve  at  any  point ;  and,  secondly,  we  can  determine  in 
what  intervals  the  curve  is  concave  upward,  in  what  concave 
downward.* 

Example.     Let  it  be  required  to  plot  the  curve 

(1)  3y  =  a-3  -  3x2  +  1. 

a)  Determine  first  its  slope  at  any  point : 

(2)  3  Dsy  =  3  aj2  -  6  x,  Dxy  =  x*-2x. 

*  There  are  two  great  applications  of  the  graphical  representation  of  a 
function.  One  is  quantitative,  the  other,  qualitative.  By  the  first  I  mean 
the  use  of  the  graph  as  a  table,  for  actual  computation.  Thus  in  the  use 
of  logarithms  it  is  desirable  to  have  a  graph  of  the  function  y  =  log10  x 
drawn  accurately  for  values  of  x  between  1  and  10  ;  for  by  means  of  such 
a  graph  the  student  can  read  off  the  logarithms  he  is  using,  correct  to  two 
or  three  significant  figures,  and  so  obtain  a  check  on  his  numerical  work. 

There  is,  however,  a  second  large  and  important  class  of  problems,  in 
which  the  character  of  a  function  is  the  important  thing,  a  minute  deter- 
mination of  its  values  being  in  general  irrelevant. 

A  case  in  point  is  the  determination  of  the  number  of  roots  of  an  alge- 
braic equation,  e.g.  x3  _  x2  _  4  x  +  l  _  0> 

Here,  we  plot  the  curve  ,        „       .         . 

r  y  =  x3  —  x%  —  4x  +  l 

and  inquire  where  it  cuts  the  axis  of  x.  For  this  purpose  it  is  altogether 
adequate  to  know  the  character  of  the  curve,  and  for  treating  this  problem 
the  methods  of  the  present  paragraph  yield  a  powerful  instrument. 


APPLICATIONS 


65 


It  is  always  useful  to  kuow  the  points  at  which  the  tangent 
to  the  curve  is  parallel  to  the  axis  of  x.  These  are  obtained 
by  setting  Dxy  —  0  and  solving.  Thus  we  get  from  (2)  the 
equation : 

The  roots  of  this  equation  are 

x  =  0         and         x  =  2 

Now  determine  accurately  the  points  having  these  abscissas, 
plot  them,  and  draw  the  tangents  there  : 


„  I      i  . 


2/U=-l- 


We  do  not  yet 
know  whether  the 
curve  lies  above  its 
tangent  in  one  of 
these  points,  or  be- 
low its  tangent ;  it 
might  even  cross 
its  tangent,  for  the 
point  might  be  a 
point  of  inflection.  These  questions  will  all  be  answered  by 
aid  of  the  second  derivative. 

b)  Compute  the  second  derivative : 

Dx*y  =  2x- 2  =2(x-l). 

We  see  that  it  is  positive  when  x  is  greater  than  1  and  nega- 
tive when  x  is  less  than  1 : 


Fig.  26 


DJy  >  0 

when 

1  <  *; 

D?y  <  0 

when 

x  <  1. 

D*y  =  0 

when 

x=  1. 

Hence  the  curve  has  a  point  of  inflection  when  x=l.     This 
is  a  most  important  point  on  the  curve.     We  will  compute  its 


66  CALCULUS 

coordinates  accurately,  determine  the  slope  of  the  curve  there, 
and  draw  accurately  the  tangent  there. 

y\z=i  =  -i;  Dzy\I=l  =  -\. 

This  is  the  last  of  the  important  tangents  which  we  need  to 
draw.  Since  the  curve  is  concave  upward  to  the  right  of  the 
line  x  =  1,  and  concave  downward  to  the  left  of  that  line,  it 
must  be  in  character  as  indicated.  We  see,  then,  that  it  cuts 
the  axis  of  x  between  0  and  1,  and  again  to  the  right  of  the 
point  x  =  1 ;  and  it  cuts  that  axis  a  third  time  to  the  left  of 
the  origin. 

These  last  two  points  can  be  located  more  accurately  by 
computing  the  function  for  a  few  simple  values  of  x. 

hence  the  curve  cuts  the  axis  of  x  between  x  =  2  and  x  =  3. 
V  r=-i  =  -  3 ; 

hence  the  curve  cuts  the  axis  between  x  =  0  and  x  =  —  1. 
Incidentally  we  have  shown  that  the  cubic  equation 

z3  -  3a;2  +  1  =  0 

has  three  real  roots,  and  we  have  located  each  between  two 
successive  integers. 

EXERCISES 

Discuss  in  a  similar  manner  the  following  curves.  In  par- 
ticular : 

a)  Determine  the  points  at  which  the  tangent  is  horizontal, 
if  such  exist,  and  draw  the  tangent  at  each  of  these  points ; 

b)  Determine  the  intervals  in  which  the  curve  is  concave 
upward,  and  those  in  which  it  is  concave  downward  ; 

c)  Determine  the  points  of  inflection,  if  any  exist,  and  draw 
the  tangent  in  each  of  these  points  ; 


APPLICATIONS  67 

d)  Draw  in  the  curve.* 

In  most  cases  it  is  desirable  to  take  2  cm.  as  the  unit. 

1.  y  =  x3  +  Sx2  -  2. 

2.  y  =  xs  —  3x  +  1. 

3.  y  —  x3  +.3x  +  1. 

4.  6 y  =  2  a'3  -  P.  a?2  -  12  x  +  6. 

5.  6</  =  2a;3  +  3x2-12.t--4. 

6.  y  =  Xs  +  x2  +  x  -+-  1. 

Suggestion.  Show  that  the  derivative  has  no  real  roots  and 
hence,  being  continuous,  never  changes  sign. 

7.  l2y  =  4:X3-6x2  +  12x-9. 

8.  y  =  2  a?2  —  x  —  x3. 

9.  12y  =  4x3  +  18x2  +  27  x  +  12. 

10.  2/  =  l-4x  +  6cc2  —  3a-3. 

11.  ?/  =  1  +  2aj-|-a;2  —  x3. 

12.  42/  =  x4  -6z2  +  8.  13.   y  =  xA  —  Sx2  +  4. 
14.      y  =  x  —  x5.  15.    ?/  =  x  +  as5. 

16.      y  =  xi  +  xi.  17.    y  =  x*  —  x2. 

18.  ;?/  =  3.T5  +  5tt3  +  15»;  +  2. 

19.  60?/  =  2a;64-15a;4  +  60a;2-30. 

6.   Relative  Maxima  and  Minima.     Points  of  Inflection.     A 

function 

(i)  y=m 

*  Since  a  curve  separates  very  slowly  from  its  tangent  near  a  point  of 
inflection,  the  material  graph  of  the  curve  must  necessarily  coincide  with 
the  material  graph  of  the  tangent  for  some  little  distance. 


68 


CALCULUS 


is  said  to  have  a  maximum  at  a  point  x  =  xQ  if  its  value  at  x0  is 
larger  than  at  any  other  point  in  the  neighborhood  of  x0.  But 
such  a  maximum  need  not  represent  the  largest  value  of  the 
function  in  the  complete  interval  a  ^  x  ^  b,  as  is  shown  by 

Fig.  27,  and  for  this  reason 
it  is  called  a  relative  maxi- 
mum, in  distinction  from 
a  maximum  maximorum, 
or  an  absolute  maximum. 

A      similar      definition 
holds  for  a  minimum,  the 
word  "  larger  "  merely  being  replaced  by  "  smaller." 

It  is  obvious  that  a  characteristic  feature  of  a  maximum  is 
that  the  tangent  there  is  parallel  to  the  axis  of  x,  the  curve 
being  concave  downward.  Similarly  for  a  minimum,  the  curve 
here  being  concave  upward.     Hence  the  following 

Test  for  a  Maximum  ok  a  Minimum.     If 

(a)  ID^I=X=0,  [£>xyUo<0, 

the  function  has  a  maximum  for  x  =  xQ;  if 

(6)  [AtfUv=o,         [Z>^]_o>0, 

it  has  a  minimum. 

The  condition  is  sufficient,  but  not  necessary  ;  cf.  §  7. 

Example.     Let  y  =  x6  —  3  x2  +  1. 

Here  Dxy  =  6  afi  -  6  x  =  6  x(x2  -  l)(a:2  +  1), 

and  hence  Dxy  =  0     for     x  =  —  1,  0, 1. 

Thus  the  necessary  condition  for  a  maximum  or  a  minimum, 
Djy  =  0,  is  satisfied  at  each  of  the  points  x  =  —  1,  0,  1. 

To  complete  the  determination,  if  possible,  compute  the 
second  derivative,         D,y  =  ^x,_^ 

and  determine  its  sign  at  each  of  these  points  : 


Fig.  28 


APPLICATIONS  69 

[D^]^.!  =  24  >  0,  .-.  x  =  —  1  gives  a  minimum; 
[Dx22/]x==o  =— 6<0,  .-.  x=  0  gives  a  maximum; 
\D}y~\x=-l  =  24  >  0,         .-.  x  =      1  gives  a  minimum. 

Points  of  Inflection.  A  point  of  inflection  is  characterized 
geometrically  by  the  phenomenon  that,  as  a  point  P  describes 
the  curve,  the  tangent  at  P 
ceases  rotating  in  the  one  di- 
rection and,  turning  back,  be- 
gins to  rotate  in  the  opposite 
direction.  Hence  the  slope 
of  the  curve,  tan  t,  has  either 
a  maximum  or  a  minimum  at 
a  point  of  inflection. 

Conversely,  if  tan  t  has  a 
maximum  or  a  minimum,  the  curve  will  have  a  point  of  inflec- 
tion. For,  suppose  tan  t  is  at  a  maximum  when  x  =  x0.  Then 
as  x,  starting  with  the  value  x0,  increases,  tan  t,  i.e.  the  slope 
of  the  curve,  decreases  algebraically,  and  so  the  curve  is  con- 
cave downward  to  the  right  of  x0.  On  the  other  hand,  as  x 
decreases,  tan  t  also  decreases,  and  so  the  curve  is  concave  up- 
ward to  the  left  of  x0. 

Now,  we  have  just  obtained  a  theorem  which  insures  us  a 
maximum  or  a  minimum  in  the  case  of  any  function  which 
satisfies  the  conditions  of  the  theorem.  If,  then,  we  choose 
as  that  function,  tan  t,  the  theorem  tells  us  that  tan  t  will 
surely  be  at  a  maximum  or  a  minimum  if 

Dx  tan  T  =  0,  DJ>  tan  T  =jfc  0. 

Hence,  remembering  that 

tan  t  =  Dxy, 
we  obtain  the  following 

Test  for  a  Point  of  Inflection.     If 

[Afrl^-o,         [Dxyuo=*o, 

the  curve  has  a  point  of  inflection  at  x=  x0. 


70 


CALCULUS 


This  test,  like  the  foregoing  for  a  maximum  or  a  minimum, 
is  sufficient,  but  not  necessary  ;  cf .  §  7. 
Example.     Let 

27 y  =  x4  +  2x3  -  12  a2  +  14 a;  -  1. 
Then  27  JD,//  =  4  Xs  +  6  x2  -  24  x  +  14, 

27  Dx*y  =  12 x2  +  12  x  -  24  =  12(aj  -  l)(x  +  2), 
27L>x3*/  =  12(2x  +  l). 

Setting  D*y  =  0,  we  get  the  points  x  =  1  and  x  =  —  2.     And 

since 

27[Z)/i/;U=  36  =£  0,  27[Dx^]^_2=-36  *  0, 

we  see  that  both  of  these  points  are  points  of  inflection. 

The   slope  of   the   curve   in   these  points   is   given  by  the 
equations 


27[A.y]*=i=0, 


27[Z)iy]x=_s=54. 


Hence  the  curve  is  parallel  to  the  axis  of  x  at  the  first  of  these 
points ;  at  the  second  its  slope  is  2. 


EXERCISES 


Test  the  following  curves  for  maxima,  minima,  and  points 
of  inflection,  and  determine  the  slope  of  the  curve  in  each 
point  of  inflection. 


1.  y  =  4x3— 15x2+12x+l. 

2.  y  =  x3  +  x4  +  x5. 


3.  6?/  =  x6-3x4+3x2-l. 

4.  y={x-l)\x+2y. 

5  —  X 

y-2+3x2' 

6.  y=(l—  x2)3. 

7.  Deduce  a  test 
for  distinguishing  be- 
tween two  such  points  of  inflection  as  those  indicated  in 
Fig.  29. 


APPLICATIONS  71 

7.  Necessary  and  Sufficient  Conditions.  In  order  to  under- 
stand the  nature  of  the  tests  obtained  in  the  foregoing  paragraph 
it  is  essential  that  the  student  have  clearly  in  mind  the  mean- 
ing of  a  necessary  condition  and  of  a  sufficient  condition.  Let 
us  illustrate  these  ideas  by  means  of  some  simple  examples. 

a)  A  necessary  condition  that  a  quadrilateral  be  a  square  is 
that  its  angles  be  right  angles.  But  the  condition  is  obviously 
not  sufficient ;  all  rectangles  also  satisfy  it. 

b)  A  sufficient  condition  that  a  quadrilateral  be  a  square  is 
that  its  angles  be  right  angles  and  each  side  be  4  in.  long. 
But  the  condition  is  obviously  not  necessary  ;  the  sides  might 
be  6  in.  long. 

c)  A  necessary  and  sufficient  condition  that  a  quadrilateral 
be  a  square  is  that  its  angles  be  right  angles  and  its  sides  be 
mutually  equal. 

As  a  further  illustration  consider  the  following.  It  is  a 
well-known  fact  about  whole  numbers  that  if  the  sum  of  the 
digits  of  a  whole  number  is  divisible  by  3,  the  number  is  divis- 
ible by  3 ;  and  conversely.  Also,  if  the  sum  of  the  digits  of  a 
whole  number  is  divisible  by  9,  the  number  is  divisible  by  9 ; 
and  conversely.     Hence  we  can  say : 

i)  A  necessary  condition  that  a  whole  number  be  divisible 
by  9  is  that  the  sum  of  its  digits  be  divisible  by  3.  But  the 
condition  is  not  sufficient. 

ii)  A  sufficient  condition  that  a  whole  number  be  divisible 
by  3  is  that  the  sum  of  its  digits  be  divisible  by  9.  But  the 
condition  is  not  necessary. 

iii)  A  necessary  and  sufficient  condition  that  a  whole  num- 
ber be  divisible  by  3  (or  9)  is  that  the  sum  of  its  digits  be 
divisible  by  3  (or  9). 

Turning  now  to  the  considerations  of  §  6,  we  see  that  a 
necessary  condition  for  a  minimum  is  that 

Dxy  =  0 


72  CALCULUS 

at  the  point  in  question,  x  =  xn.  But  this  condition  is  not 
sufficient.  When  it  is  fulfilled,  the  function  may  have  a 
maximum,  or  it  may  have  a  point  of  inflection  with  horizontal 
tangent. 

On  the  other  hand,  the  condition 

[A2TU,  =  0,  [Z^]x=Xo>0 

is  sufficient  for  a  minimum.  But  it  is  not  necessary.  Thus 
the  function 

(1)  y=a* 

obviously  has  a  minimum  when  x  =  0.  The  necessary  condi- 
tion, Dxy  =  0,  is  of  course  fulfilled  : 

But  here       Dx2y  =  12  x\  and  [£>,2#]x=o 

is  not  positive ;  it  is  0. 

Again,  as  was  shown  in  §  4,  a  necessary  condition  for  a  point 

of  inflection  is  that  _  n 

Dxhj  =  0 

at  that  point.  But  this  condition  is  not  sufficient.  Thus  in 
the  case  of  the  curve  (1)  this  condition  is  fulfilled  at  the 
origin.     But  the  origin  is  not  a  point  of  inflection. 

Remark.  It  may  seem  to  the  student  that  such  tests  are 
unsatisfactory  since  the}"  do  not  apply  to  all  cases  and  thus 
appear  to  be  incomplete.  But  their  very  strength  lies  in  the 
fact  that  they  do  not  tell  the  truth  in  too  much  detail.  They 
single  out  the  big  thing  in  the  cases  which  arise  in  practice 
and  yield  criteria  which  can  be  applied  with  ease  to  the  great 
majority  of  these  cases. 

8.  Velocity ;  Rates.  By  the  average  velocity  with  which  a 
point  moves  for  a  given  length  of  time  t  is  meant  the  distance 
5  traversed  divided  by  the  time  : 

average  velocity  =-• 

t 


APPLICATIONS  73 

Thus  a  railroad  train  which  covers  the  distance  between  two 
stations  15  miles  apart  in  half  an  hour  has  an  average  speed 
of  15/|=  30  miles  an  hour. 

When,  however,  the  point  in  question  is  moving  sometimes 
fast  and  sometimes  slowly,  we  can  describe  its  speed  approxi- 
mately at  any  given  instant  by  considering  a  short  interval 
of  time  immediately  succeeding  the  instant  t0  in  question,  and 
taking  the  average  velocity  for  this  short  interval. 

For  example,  a  stone  dropped  from  rest  falls  according  to 
the  law :  .  _  „ 

To  find  how  fast  it  is  going  after  the  lapse  of  t0  seconds.     Here 

(1)  s0=16t0*. 

A  little  later,  at  the  end  of  t'  seconds  from  the  beginning  of 
the  fall, 

(2)  s'  =  16t'2 

and  the  average  velocity  for  the  interval  of  t'  —  t0  seconds  is 

(3)  s'  -So  ft<  per  seCond. 

Let  us  consider  this  average  velocity,  in  particular,  after  the 

lapse  of  1  second : 

t0  =  1,  s0  =  16. 

Let  the  interval  of  time,  t'  —  t0,  be  tl  sec.     Then 

s'  =  16  x  1.1"  =  19.36, 

y--sg  =  3^36  =  33<6  ft  a  second 
V  -  to        .1 

Thus  the  average  velocity  for  one-tenth  of  a  second  immedi- 
ately succeeding  the  end  of  the  first  second  of  fall  is  33.6  ft.  a 
second. 

Next,  let  the  interval  of  time  be  j^  sec.  Then  a  similar 
computation  gives,  to  three  significant  figures  : 

s'  ~  s°  =  32.2  ft.  a  second. 

t'-to 


74  CALCULUS 

And  when  the  interval  is  taken  as  ^00  sec.,  the  average 
velocity  is  32.0  ft.  a  second. 

These  numerical  results  indicate  that  we  can  get  at  the 
speed  of  the  stone  at  any  desired  instant  to  any  desired  degree 
of  accuracy  by  direct  computation ;  we  need  only  to  reckon 
out  the  average  velocity  for  a  sufficiently  short  interval  of 
time  succeeding  the  instant  in  question. 

We  can  proceed  in  a  similar  manner  when  a  point  moves 
according  to  any  given  law.  Can  we  not,  however,  by  the  aid 
of  the  Calculus  avoid  the  labor  of  the  computations  and  at  the 
same  time  make  precise  exactly  what  is  meant  by  the  velocity 
of  the  point  at  a  given  instant?  If  we  regard  the  interval 
of  time  t'  — 10  as  an  increment  of  the  variable  t  and  write 
t'  —  tQ  =  At,  then  s'  —  s0  =  As  will  represent  the  corresponding 
increment  in  the  function,  and  thus  we  have  : 

average  velocity  =      • 
At 

Now  allow  At  to  approach  0  as  its  limit.  Then  the  average 
velocity  will  in  general  approach  a  limit,  and  this  limit  we  take 
us  the  definition  of  the  velocity,  v,  at  the  instant  (tt : 

lim  (average  velocity  from  t  =  t0  to  t  =  t') 

=  actual  velocity  *  at  instant  t  =  t0, 

or  v  =  lim  —  =  Dts. 

A«=y)  At 

Hence  it  appears  that  the  velocity  of  a  point  is  the  time- 
derivative  of  the  space  it  has  traveled.  In  the  case  of  a 
freely  falling  body  this  velocity  is 

v  =  Dts  =  32 1. 

In  the  foregoing  definition,  s  has  been  taken  as  the  distance 
actually  traversed  by  the  moving  point,  P.  More  generally, 
let  s  denote  the  length  of  the  arc  of  the  curve  on  which  P  is 
moving,  s  being  measured  from  an  arbitrarily  chosen  fixed 

*  Sometimes  called  the  instantaneous  velocity. 


APPLICATIONS  75 

point  of  that  curve.  Either  direction  along  the  curve  may  be 
chosen  as  the  positive  sense  for  s.  Thus,  in  the  case  of  a 
freely  falling  body,  s  might  be  taken  as  the  distance  of  the 
body  above  the  ground.  If  h  denotes  the  initial  distance,  then 
s  +  s'  =  h,  A 

where  s'  denotes  the  distance  actually  traversed  by  P 
at  any  given  instant.     Hence 

Dts  +  Dts'  =  0,  s" 


or  Dts  =  —  Dts'.  Fig.  30 

Here  Dts  gives  numerically  the  value  of  the  velocity,  but  Dts  is 
a  negative  quantity. 

We  will,  accordingly,  extend    the  conception   of   velocity, 
defining  the  velocity  v  of  the  point  as  Dts: 

v  =  Dts. 

Thus  the  numerical  value  of  v  or  Dts  will  always  give  the 
speed,  or  the  value  of  the  velocity  in  the  earlier  sense.  In 
case  s  increases  with  the  time,  Dts  is  positive  and  represents 
the  speed.  If,  however,  s  decreases  with  the  time,  Dts  is  nega- 
tive, and  the  velocity,  v,  is  therefore  here  negative,  the  speed 
now  being  given  by  —  v  or  —  Dts.     In  all  cases, 

Speed=|y|  =\Dts\. 

Example.     Let  a  body  be  projected  upward  with  an  initial 
velocity  of  96  ft.  a  second.     Assuming  from  Physics  the  law 

that  s  =  96t-l6t2, 

find  its  velocity        a)  at  the  end  of  2  sec. 

b)  at  the  end  of  5  sec. 

Solution.     By  definition,  the  velocity  at  any  instant  is 

v  =  D.s  =  96  -S2t. 
Hence 

a)  v|fc=2=64. 

b)  v  |e=5  =  -  64. 


76  CALCULUS 

The  meaning  of  these  results  is  that,  at  each  of  the  two 
instants,  the  speed  is  the  same,  namely,  64  ft.  a  second  (and 
the  height  above  the  ground  is  also  seen  to  be  the  same, 
s  =  128  ft.).  But  when  t  =  2,  Dts  is  positive ;  hence  s  is  in- 
creasing with  the  time  and  the  body  is  rising.  When  t  =  o, 
Dts  is  negative ;  hence  s  is  decreasing  with  the  time,  and  the 
body  is  descending. 

Rates.  Consider  any  length  or  distance,  r,  which  is  chang- 
ing with  the  time,  and  so  is  a  function  of  the  time.  Let  r0 
denote  the  value  of  r  at  a  given  instant,  t  =  t0,  and  let  r'  be  the 
value  of  r  at  a  later  instant,  t  =  t'.  Then  the  increase  in  r 
will  be  r'  —  r0  =  Ar  and  that  in  t  will  be  t'  —  1 0  =  At.  Thus  in 
the  interval  of  time  of  At  seconds  succeeding  the  instant  t  =  t0, 

average  rate  of  increase  of  r  = 

A^ 

Now  let  At  approach  0  as  its  limit.  Then  the  average  rate  of 
increase  will  in  general  approach  a  limit,  and  this  limit  ice  take 
as  the  definition  of  the  rate  of  increase  of  r  at  the  instant  t0: 

lim  (average  rate  of  increase  from  t  =  t0  to  t  =  t') 

=  actual  rate  of  increase  at  instant  t  =  t0 

r      Ar       ^ 
=  Inn  —  =  Dtr. 

Af  =  »  At 

In  other  words,  the  rate  at  ivhich  r  is  increasing  at  any  in- 
stant is  defined  as  the  time-derivative  of  r. 

If  r  is  decreasing,  Dtr  will  be  a  negative  quantity  ;  and  con- 
versely, if  Dtr  is  negative,  then  r  is  decreasing.  In  either  case, 
the  numerical  value  of  Dtr  gives  the  rate  of  change  of  r ;  just 
as,  in  the  case  of  velocities,  the  numerical  value  of  Dts  gives 
the  speed. 

More  generally,  instead  of  r,  we  may  have  any  physical 
quantity,  u,  as  an  area  or  a  volume  or  the  current  in  an  electric 
circuit  or  the  number  of  calories  in  a  given  body. 


APPLICATIONS 


77 


In  all  these  cases,  the  rate  at  which  u  is  increasing  is  defined 
as  the  time-derivative  of  u,  i.e.  as  Dtu ;  and  the  rate  of  change 
of  u  is  |  Dtu  \. 

Example.  At  noon,  one  ship  is  steaming  east  at  the  rate  of 
18  miles  an  hour,  and  a  second  ship,  40  miles  north  of  the  first, 
is  steaming  south  at  the  rate  of  20  miles  an  hour.  At  what 
rate  are  they  separating  from  each  other  at  one  o'clock  ? 

Solution.     The  relation  between  r  and   t  is 
here  given  by  the  Pythagorean  Theorem  : 

r2  =  (40-20*)2+(18*)2, 


or 

(1) 
Hence 

(2) 


r2  =  1600  -  1600*  +  724*2. 


r  =  V1600  -  1600*  +  724*2 


Fig.  31 


We  wish  to  find  D-tr.  This  can  be  done  by  differentiating 
equation  (2)  ;  but  that  would  be  poor  technique,  since  it  is 
simpler  to  differentiate  equation  (1)  through  with  respect  to  t: 


(3) 


2rDtr  = 


1600  +  1448*, 
800  +  724* 


Equation  (3)  gives  the  rate  at  which  r  is  increasing  at  any 
instant  t ;  i.e.  t  hours  past  noon,  or  at  t  o'clock. 
Setting  now,  in  particular,  t  =  1,  we  obtain : 


D.r 


76 


V724 


=  -2.825. 


The  meaning  of  this  result  is  twofold.  First,  since  Dtr  is 
negative  when  t  =  1,  the  ships  are  not  receding  from  each 
other,  but  are  coming  nearer  together.  Secondly,  the  rate  of 
change  of  the  distance  between  them  is,  at  one  o'clock,  2.825 
miles  an  hour. 


78  CALCULUS 

Let  the  student  determine  how  long  they  will  continue  to 
approach  each  other,  and  what  the  shortest  distance  between 
them  will  be. 

Remark.  It  is  important  for  the  student  to  reflect  on  the 
method  of  solution  of  this  problem,  since  it  is  typical.  We 
were  asked  to  find  the  rate  of  recession  at  just  one  instant,  t  =  l. 
We  began  by  determining  the  rate  of  recession  generally,  i.e. 
for  an  arbitrary  instant,  t  —  t.  Having  solved  the  general 
problem,  we  then,  as  the  last  step  in  the  process,  brought  into 
play  the  specific  value  of  t  which  alone  we  cared  for,  namely, 
t  =  l. 

The  student  will  meet  this  method  again  and  again,  —  in 
integration,  in  mechanics,  ha  series,  etc.  We  can  formulate 
the  foregoing  remark  suggestively  as  follows  :  By  means  of 
the.  Calculus  we  can  often  determine  a  particular  physical 
quantity,  like  a  velocity,  an  area,  or  the  time  it  takes  a  body, 
acted  on  by  known  forces,  to  reach  a  certain  position.  The 
method  consists  in  first  determining  a  function,  whereby  the 
general  problem  is  solved  for  the  variable  case ;  and  then,  as 
the  last  step  in  the  process,  the  special  numerical  values  with 
which  alone  the  proposed  question  is  concerned,  are  brought 
into  play. 

EXERCISES 

1.  The  height  of  a  stone  thrown  vertically  upward  is  given 

by  the  formula :  .  _        .  _  „ 

J  i       s=ASt  —  Hit-. 

When  it  has  been  rising  for  one  second,  find  (a)  its  average 
velocity  for  the  next  ^  sec. ;  (b)  for  the  next  y^j  sec.  ;  (c)  its 
actual  velocity  at  the  end  of  the  first  second ;  (d)  how  high  it 
will  rise. 

Ans.  (a)  14.4  ft.  a  second ;  (6)  15.84  ft.  a  second  ;  (c)  16  ft. 
a  second ;  (d)  36  ft. 

2.  One  ship  is  80  miles  due  south  of  another  ship  at  noon, 
and  is  sailing  north  at  the  rate  of  10  miles  an  hour.     The 


APPLICATIONS  79 

second  ship  sails  west  at  the  rate  of  12  miles  an  hour.  Will 
the  ships  be  approaching  each  other  or  receding  from  each 
other  at  2  o'clock  ?  What  will  be  the  rate  at  which  the  dis- 
tance between  them  is  changing  at  that  time  ?  How  long  will 
they  continue  to  approach  each  other  ? 

3.  If  two  ships  start  abreast  half  a  mile  apart  and  sail  due 
north  at  the  rates  of  9  miles  an  hour  and  12  miles  an  hour, 
how  far  apart  will  they  be  at  the  end  of  half  an  hour  ?  How 
fast  will  they  be  receding  at  that  time  ? 

4.  Two  ships  are  steaming  east,  one  at  the  rate  of  18  miles 
an  hour,  the  other  at  the  rate  of  24  miles  an  hour.  At  noon, 
one  is  50  miles  south  of  the  other.  How  fast  are  they  sepa- 
rating at  7  p.m.  ? 

5.  A  ladder  20  ft.  long  rests  against  a  house.  A  man 
takes  hold  of  the  lower  end  of  the  ladder  and  walks  off  with 
it  at  the  uniform  rate  of  2  ft.  a  second.  How  fast  is  the  upper 
end  of  the  ladder  coming  down  the  wall  when  the  man  is  4  ft. 
from  the  house  ? 

6.  A  kite  is  150  ft.  high  and  there  are  250  ft.  of  cord  out. 
If  the  kite  moves  horizontally  at  the  rate  of  4  m.  an  hour 
directly  away  from  the  person  who  is  flying  it,  how  fast  is  the 
cord  being  paid  out  ?  Ans.    31  m.  an  hour. 

7.  A  stone  is  dropped  into  a  placid  pond  and  sends  out 
a  series  of  concentric  circular  ripples.  If  the  radius  of  the 
outer  ripple  increases  steadily  at  the  rate  of  6  ft.  a  second, 
how  rapidly  is  the  area  of  the  water  disturbed  increasing  at 
the  end  of  2  sec.  ?  Ans.    452  sq.  ft.  a  second. 

8.  A  spherical  raindrop  is  gathering  moisture  at  such  a 
rate  that  the  radius  is  steadily  increasing  at  the  rate  of  1  mm. 
a  minute.  How  fast  is  the  volume  of  the  drop  increasing 
when  the  diameter  is  2  mm.  ? 

9.  A  man  is  walking  over  a  bridge  at  the  rate  of  4  miles  an 
hour,  and  a  boat  passes  under  the  bridge  immediately  below 
him  rowing  8  miles  an  hour.     The  bridge  is  20  ft.  above  the 


80  CALCULUS 

boat.     How  fast  are  the  boat  and  the  man  separating  3  min- 
utes later  ? 

Suggestion.  The  student  should  make  a  space  model  for 
this  problem  by  means,  for  example,  of  the  edge  of  a  table,  a 

crack  in  the  floor,  and  a  string  ;  or 
by  two  edges  of  the  room  which  do 
not  intersect,  and  a  string.  He 
should  then  make  a  drawing  of  his 
model  such  as  is  here  indicated. 

10.    A     locomotive     running    30 
miles  an  hour  over  a  high   bridge 
dislodges   a   stone   lying   near    the 
track.     The  stone  begins  to  fall  just  as  the  locomotive  passes 
the  point  where  it  lay.     How  fast  are  the  stone  and  the  loco- 
motive separating  2  sec.  later  ?  * 

11.  Solve  the  same  problem  if  the  stone  drops  from  a  point 
40  ft.  from  the  track  and  at  the  same  level,  when  the  locomo- 
tive passes. 

12.  A  lamp-post  is  distant  10  ft.  from  a  street  crossing  and 
GO  ft.  from  the  houses  on  the  opposite  side  of  the  street.  A 
man  crosses  the  street,  walking  on  the  crossing  at  the  rate  of 
4  miles  an  hour.  How  fast  is  his  shadow  moving  along  the 
walls  of  the  houses  when  he  is  halfway  over? 

*  Bocher,  Plane  Analytic  Geometry,  p.  230. 


CHAPTER   IV 
INFINITESIMALS    AND    DIFFERENTIALS 

1.  Infinitesimals.  An  infinitesimal  is  a  variable  which  it  is 
desirable  to  consider  only  for  values  numerically  small  and 
which,  when  the  formulation  of  the  problem  in  hand  has  pro- 
gressed to  a  certain  stage,  is  allowed  to  approach  0  as  its  limit. 

Thus  in  the  problem  of  differentiation,  or  finding  the  limit 

(1)  lim^=Z^, 

A*=y>  Ax 

Ax  and  Ay  are  infinitesimals ;  for  we  allow  Ax  to  approach  0 
as  its  limit,  and  then  Ay  also  approaches  0.. 

Again,  if  we  denote  the  value  of  the  difference  Ay/ Ax  —  Dzy 
by  e,  so  that 

(2)  ^-D,y  =  t, 

Ax 

then  e  is  an  infinitesimal.  For,  when  Ax  approaches  0,  the 
left-hand  side  of  equation  (2)  approaches  0,  and  so  e  is  a  vari- 
able which  approaches  0  as  its  limit,  i.e.  an  infinitesimal. 

Principal  Infinitesimal.  When  we  are  dealing  with  a  num- 
ber of  infinitesimals,  «,  /?,  y,  etc.,  it  is  usually  possible  to 
choose  any  one  of  them  as  the  independent  variable,  the  others 
then  becoming  functions  of  it,  or  dependent  variables.  That 
infinitesimal  which  is  chosen  as  the  independent  variable  is 
called  the  principal  infinitesimal. 

Thus,  if  the  infinitesimals  are  «  and  /3,  and  if 

(3)  /3  = 


l+3a 

81 


82  CALCULUS 

it  is  natural  to  choose  a  as  the  principal  infinitesimal.  But 
it  is  perfectly  possible  to  take  /?  as  the  principal  infinitesimal. 
a  then  becomes  the  dependent  variable,  and  is  expressed  in 
terms  of  (3  by  solving  equation  (3)  for  a : 


(■*) 


2-3/? 


Order  of  Infinitesimals.  We  are  going  to  separate  infinitesi- 
mals into  classes,  according  to  the  relative  speed  with  which 
they  approach  0.  Suppose  we  let  a  set  the  pace,  taking  on 
the  values  .5,  .1,  .01,  .001,  etc.  Consider,  for  example,  a2. 
Then  a2  takes  on  the  respective  values  .25,  .01,  .0001,  etc.,  and 
hence  runs  far  ahead  of  a : 


a 

.5 

.1 

.01 

.001  ••• 

a- 

.25 

.01 

.0001 

.000001  ... 

Furthermore,  the  closer  the  two  get  to  0,  the  relatively  nearer 
a2  is  to  0.  Thus,  when  a  =  .5,  a2  is  twice  as  close  ;  but  when 
a  =  .01,   a2  is  one  hundred  times  as  close ;  and  so  on. 

Again,  consider  the  infinitesimal  \a.  It  is  always  twice  as 
close  to  0  as  a  is.  Similarly,  10  a  is  always  one-tenth  as  close 
as  a. 

From  these  examples  we  see  that  there  is  a  decided  difference 
between  the  relative  behavior  of  a  and  ka  on  the  one  hand, 
and  that  of  a  and  a2  on  the  other.  For,  ka  is  keeping  pace 
relatively  with  a,  whereas  a2  runs  indefinitely  ahead  of  a,  rela- 
tively. Consequently,  we  should  put  ka  into  the  same  class 
with  a,  whereas  a2  forms  the  starting  point  for  a  new  class. 
To  this  latter  class  would  belong  such  infinitesimals  as  ^a2  or 
4a2  —  az ;  and  the  former  class  would  include,  for  example, 
2a  +  3a-  and  -f^a  —  1000a8.  Let  the  student  make  out  a 
table  like  the  above  for  each  of  these  examples. 

What  is  the  common  property  of  all  infinitesimals  of  the 
same  class  ?  Is  it  not,  that,  for  two  infinitesimals,  the  relative 
speed  with  which  they  approach  0  is  nearly,  or  quite,  a  fixed 
number  not  zero  ?     It  is  this  idea  which  lies  at  the  bottom  of 


INFINITESIMALS   AND    DIFFERENTIALS  83 

the  conception  of  the  order  of  an  infinitesimal,  and  it  is  for- 
mulated in  a  precise  definition  as  follows  : 

Definition.     Two  infinitesimals,  /?  and  y,  are  said  to  be  of 
the  same  order  if  their  ratio  approaches  a  limit  not  0 ; 

lim^=  A>0. 

y 

Thus  £  =  2  a  +  «2         and         y  =  3  a  -  a3 

are  of  the  same  order.     For, 

P  __2  a  +  a2      2  +  « 
y      3  a  —  a3      3  —  a2 

and  hence,  when  «  approaches  0, 

..     /3      v     2  +  «      2      . 

lim^=lim-^--  =  -^0. 

y  3  —  a2     3 

Similarly,  12  a2  +  3  a5  and  6  a2  —  7  a3  are  infinitesimals  of  the 
same  order. 

An  infinitesimal  f3  is  said  to  be  of  higher  order  than  y  if 

lim  £  =  0, 

y 

Thus  if  j8  =  9«2         and         y  =  2 a  +  5a\ 

fi  is  of  higher  order  than  a.     For, 

£_       9<*2  9« 

y      2«  +  5«4     2  + 5a3' 

and  hence,  when  a  approaches  0, 

lim  £  =  lim— ^ :=0. 
y  2  +  5  a3 

Finally,  /?  is  said  to  be  of  lower  order  than  y  if 
(5)  lim  —  =  oo  , 

y 

(read  :  "/3/y  becomes  infinite  ";  not  "  /?/y  equals  infinity."  *). 

*  The  student  should  now  turn  back  to  Chapter  II,  §  5,  and  read  again 
carefully  what  is  said  there  about  infinity.     In  particular,  he  should  iui- 


84  CALCULUS 

Thus  if  ($  =  Va         and         y  =  6  a  +  a3, 

/?  is  of  lower  order  than  y.     For 

ff_     V^     =  1 

When  a  approaches  0,  it  is  evident  that  the  last  fraction  in- 
creases without  limit,  or  n 

liin  °  =  oo  ■ 

y 

.FtVs£  Order,  Second  Order,  etc.  An  infinitesimal  ^  is  said  to 
be  of  the  Jirst  order  if  it  is  of  the  same  order  as  the  principal 
infinitesimal,  a  ;  i.e.  if  ^ 

lim"  =  A'^0. 
it 

If  /?  is  of  the  same  order  as  a2,  i.e.  if 

lim4  =  A^0, 
or 

then  (3  is  said  to  be  of  the  second  order.  And,  generally,  if  /3 
is  of  the  same  order  as  a",  i.e.  if 

lim-£=Jg-=£0, 
a" 

then  /3  is  said  to  be  of  the  n-th  order. 
Thus  if 

£  =  2«       or       0  =  _ ?_       or       £  =  «  +  a2, 
2  —  a 
then  /?  is  of  the  first  order. 

But  if 

|3  =  2«2+«3       or       j8  =  — —       or       0  =  a2, 

3  +  a 

then  /?  is  of  the  second  order. 

press  on  his  mind  the  fact  that  infinity  is  not  a  limit  and  that  in  the 
notation  used  in  (5)  the  =  sign  does  not  mean  that  one  number  is  equal 
to  another  number.  The  formula  is  not  an  equation  in  the  sense  in  which 
2x  =  3  or  a2  —  b2  =  (a  —  b) (a  +  b)  is  an  equation.  The  formula  means 
no  more  and  no  less  than  that  the  variable  p/y  increases  in  value  without 
limit. 


INFINITESIMALS  AND   DIFFERENTIALS  85 

Ifj8=Va,  then  p  _  1 

i  —  L> 

a2 

and  lim  -£  =  1  =£  0. 

a2 
Hence  j3  is  of  the  order  i. 

It  is  easily  seen  that  if  two  infinitesimals  /8  and  y  are,  under 
the  present  definition,  each  of  order  n,  then  they  also  satisfy 
the  earlier  definition  of  being  of  the  same  order.     For,  let 

lim  -£  =  A>  0         and         lim  -£  =^  X  =£  0. 
a"  a" 

Then,  if  we  denote  the   differences   ft/an  —  K  and   y/a"  —  L 
respectively  by  e  and  rj,  so  that 

(6)  L-K=t.        and         2--L  =  V) 

«"  a" 

these  variables,  e  and  77,  will  be  infinitesimals.     For,  the  left- 
hand  side  of  each  of  the  equations  (6)  approaches  0. 
From  equations  (6)  it  follows  that 

£  =  K+  e        and         2-  =  Z  +  v. 

a"  a" 

On  dividing  one  of  these  equations  by  the  other  we  have : 

y      L  +  rj 
We  are  now  ready  to  allow  a  to  approach  0  as  its  limit.     Then 

lim£=lim^±f. 

y  L  +  r] 

By  Theorem  III  of  Chapter  2,  §  5  this  last  limit  has  the  value 

lim  g+*  =  lim  (*"+«)-*'■ 
L  +  rj     lim  (L  +  rj)      L 

Hence,  finally  Um  £  =  A'^  Q>  q  e  d 

y        Z, 


86  CALCULUS 

EXERCISES 

1.  Show  that 

B  =  5u  —  11«2  +  «3         and         y  =  7a  +  a4 
are  infinitesimals  of  the  same  order. 

2.  Show  that 

/8  =  2a-3«2        and         y  =  2«  +  «4 

are  infinitesimals  of  the  same  order,  but  that  their  difference, 
B  —  y,  is  of  higher  order  than   B  (or  y). 

3.  Show  that  B  = is  an  infinitesimal  of  the  second 

H      a3 -2 

order,  referred  to  a  as  principal  infinitesimal. 

4.  Show  that  /8  =  Va2-f-U«5  is  of  the  first  order,  referred 
to  a. 


5.  Show  that  B  =  V2a+  13a3  is  of  lower  order  than  a. 

6.  Show  that  the  order  of  (3  in  question  5  is  n  =  i. 

Determine  the  order  of  each  of  the  following  infinitesimals, 
referred  to  a  as  the  principal  infinitesimal : 


7. 

\a  +  18  «3. 

11. 
12. 
13. 

14. 

's/a3  —  a. 

-«+V2«3 

7  a2 
13 -« 

-f-«4. 

8. 

ty-a12  +  a13. 

y. 

V2a2  —  tc3. 

10. 

/2a2  +  «5 

4/3«6  +  4a4t 

*  »  -  7  «  ^    a2  +  2 

15.  If  /?  and  y  are  infinitesimals  of  orders  n  and  m  respec- 
tively, show  that  their  product,  By,  is  an  infinitesimal  of  order 
n  +  m. 

16.  If  B  and  y  are  infinitesimals  of  the  same  order,  show 
that  their  sum  is,  in  general,  an  infinitesimal  of  the  same 
order. 

Are  there  exceptions  ?     Illustrate  by  examples. 


INFINITESIMALS   AND    DIFFERENTIALS  87 

2.  Continuation ;  Fundamental  Theorem.  Principal  Part  of 
an  Infinitesimal.  Let  fi  be  an  infinitesimal  of  order  n,  and 
let  a  be  the  principal  infinitesimal.     Then 

lim-£=iT=£0. 

an 

Moreover,  as  pointed  out  in  the  last  paragraph, 

(1)  \=K+^ 

an 

where  e  is  infinitesimal.     From  (1)  it  follows  that 

(2)  p  =  Ka'1  +  €u\ 

This  last  equation  gives  a  most  important  analysis  (i.e.  break- 
ing up)  of  (3  into  two  parts,  each  of  which  is  simple  for  its 
own  peculiar  reason. 

i)  Ka"  is  the  simplest  infinitesimal  of  the  nth  order  imagi- 
nable,—  a  monomial  in  the  independent  variable,  the  function 

?/  =  Kxr. 

ii)  en'1  is  an  infinitesimal  of  higher  order  than  the  rath. 
The  first  part,  Ka",  is  called  the  principal  part  of  (3. 
By  far  the  most  important  case  in  practice  is  that  of  infini- 
tesimals (3  of  the  first  order,  n  =  1.     Here 

a"      a 
and  p  =  Ka  +  ta. 

Hence  we  see  that  the  principal  part  of  an  infinitesimal  of  the 
first  order  is  proportional  to  the  principal  infinitesimal. 

Example  1.     Let  ft  =  la  —  a2. 

Then  /3  is  obviously  of  the  first  order,  or  n  =  1,  and  here 

S.-Z-2-a. 

an      a 


88  CALCULUS 

Clearly,  then,  K=2,         e  =  —  a, 

and  the  principal  part  of  /3  is  2a. 

Example  2.     Let  R  _     2  a1 


7-4« 
Here,  obviously,  n  =  2,  and 

v     $      r  2  2 

Inn  —  =  lim =  -• 

a2  7  —  4  a      7 

2 
Hence  K=—     By  definition, 

In  the  present  case,  then, 

9  9 


7  —  4  a      7      7  (7  —  4  a) 

EXERCISE 

Determine  the  principal  parts  of  a  goodly  number  of  the 
infinitesimals  occurring  in  the  Exercises  at  the  end  of  §  1. 

Equivalent  Infinitesimals.     Two  infinitesimals,  as  j3  and  y, 
shall  be  said  to  be  equivalent  if  the  limit  of  their  ratio  is  unity  : 

lim£=l. 

y 

For  example,  the  following  pairs  of  infinitesimals  are  equiv- 
alent: • 

i)  2  a  -f  a-  and         2  a  +  a?  ; 

ii)  \  a2  —  a3  and         4-  a2  +  a3  ; 

iii)  V2a  +  5a':       and  V2a  —  7  a4. 

An  infinitesimal  and  its  principal  part  are  always  equivalent 
infinitesimals.     For,  if  Kan  is  the  principal  part  of  /?,  then 

(3  =  Kan  +  t;, 


INFINITESIMALS   AND    DIFFERENTIALS  89 

where  rj  is  of  higher  order  than  Kan.     Hence 

-£-=l+^L,  lim-f-  =  l  +  lim  -2-. 

Kan  Jut'1  Kan  Kan 

But  lim  rj/Kan  =  0,  and  the  statement  is  established. 

Two  infinitesimals  which  have  the  same  principal  parts  are 
equivalent,  and  conversely. 

Equivalent  infinitesimals  are  of  the  same  order ;  but  the 
converse  is  not  true. 

The  difference  between  two  equivalent  infinitesimals,  (3  and 
y,  namely,  /5  —  y,  is  of  higher  order  than  ft  or  y.     For 

y       y 

hence  lim  "  ~  ^  =  lim  (  "  —  1 

y  \y 

lim^-l  =  0,  q.  e.d. 

Conversely,  if  /?  and  y  are  two  infinitesimals  whose  differ- 
ence, (3  —  y,  is  of  higher  order  than  /3  or  y,  then  /3  and  y  are 
equivalent. 

For,  since  fLu  =  H-ly 

y       y 

it  follows  that  linY£_;A=lim£^. 

The  right-hand  side  of  this  equation  is  0  by  hypothesis,  and 
the  left-hand  side  is  equal  to 

lim^-1. 

y) 

Hence  lim  "  =  1,  q.  e.  d. 

y 

We  come  now  to  a  theorem  of  prime  importance  in  the 
Infinitesimal  Calculus. 


90  CALCULUS 

Fundamental  Theorem.  The  limit  of  the  ratio  of  two  infini- 
tesimals, „ 

lime, 

y 

is  unchanged  if  the  numerator  infinitesimal  (3  be  replaced  by  any 
equivalent  infinitesimal  /?'  and  the  denominator  infinitesimal  y  be 
replaced  by  any  equivalent  infinitesimal  y'. 
In  other  tvords :  „  „, 

lim  P  =  lim  &- 

y  y 

provided      lim£=1  and  lim_y_=L 

P  y 

The  proof  is  immediate.     It  is  obvious  that 

/?'_£7?y 

y     Pri 

Hence  by  Theorem  II,  Chapter  II,  §  5  we  have 
lim  £  =  Aim  £Ylim£  Vlim  *A 

But  the  first  and  third  limits  on  the  right-baud  side  are  each 
equal  to  1  by  hypothesis.     Hence 

lim"  =  lim",  q.  e.  d. 

y  y 

The  theorem  can  be  stated  iu  the  following  equivalent 
form : 

The  limit  of  the  ratio  of  two  infinitesimals  is  the  same  as  the 
limit  of  the  ratio  of  their  principal  parts. 

The  student  must  not  generalize  from  this  theorem  aud 
infer  that  an  infinitesimal  can  always  aud  for  all  purposes  be 
replaced  by  an  equivalent  infinitesimal.     Thus  if 

/3  =  2a  +  a3        and         y  =  2« -  a2, 

their  difference,  (3  —  y  =  a3  +  a2, 


INFINITESIMALS   AND    DIFFERENTIALS  91 

is  an  infinitesimal  of  the  second  order.     On  the  other  hand, 

y'  =  2« 

is  equivalent  to  y.  But  it  is  not  true  that  the  difference  of  (3 
and  y',  namely,  ^  _      =  ^ 

is  an  infinitesimal  of  the  second  order.  It  is  obviously  of 
order  3.  Thus  replacing  y  by  an  equivalent  infinitesimal  has 
here  changed  the  order  of  the  difference  ft  —  y. 

3.   Differentials.     Let      y=f(x) 

be  a  function  of  x,  and  let  D,y  be  its  derivative : 

lim  — "  =  Dxy. 
A*=y>  Ax 

Let  the  difference  Ay  /Ax  —  Dxy  be  denoted  by  c.     Then 

and 

(1)  Ay  =  Z)^  Aoj  +  e  A.t. 

Since  a;  is  the  independent  variable,  Ax  can  be  taken  as  the 
principal  infinitesimal.  Djy  does  not  vary  with  Ax;  it  is  a 
constant,  for  we  are  considering  its  value  at  a  fixed  point 
x  =  x0.     Since,  moreover,  Dxy  is  not  in  general  zero,  equation 

(1)  represents  Ay  as  the  sum  of  its  principal  part,  DxyAx,  and 
an  infinitesimal  of  higher  order,  cAx. 

Definition  of  a  Differential.  The  expression  Dxy  Ax  is  called 
the  differential  of  the  function,  and  is  denoted  by  dy : 

(2)  dy  =  DxyAx,         or         df{x)=DJ(x)Ax. 

(read  :  "  differential  y  "  or  "  differential  f(x) "  or  "  dy"  etc.). 

Thus  if  y  =  x2, 

dy  =  2xAx,        or        dx2  =  2xAx. 


92 


CALCULUS 


Since  the  definition  (2)  holds  for  every  function  y  =/(x),  it 
can  be  applied  to  the  particular  function 

f(x)=x, 


dx  =  Djc  Ax  =  A.'-. 


Hence 

(3) 

But  it  is  not  in  general  true  that  Ay  and  dy  are  equal,  since  e 
is  in  general  different  from  0.  Thus  we  see  that  the  differen- 
tial of  the  independent  variable  is  equal  to  the  increment  of  that 
variable;  but  the  differential  of  the  dependent  variable  is  not  in 
general  equal  to  the  increment  of  that  variable. 

By  means  of  (3)  equation  (2)  can  now  be  written  in  the  form 

(4)  dy  =  Dxydx. 

Hence 


(5) 


ax 


Geometrically,  the  increment  A?/  of  the  function  is  repre- 
sented by  the  line  MP',  Fig.  33  ;    and  the  differential,  dy,  is 

equal  to  MQ,  for  from  (5) 

dy 

tan  t  =  — 

dx 

or        dy  =  dx  tan  t. 

In  other  words,  Ay  repre- 
sents the  distance  from  the 
level  of   P  to  the    cur 
when  x  =  x' ;  dy,  the  dis- 
tance from  the  level  of  P 


to  the  tangent. 


Moreover,  the  difference 
Ay  —  dy  =  eAx 


is  shown  geometrically  as  the  line  QP',  and  is  obviously  from 
the  figure  an  infinitesimal  of  higher  order  than  Ax  =  PM. 

It  is  also  clear  from  the  figure  that  Ay  and  dy  are  equal 
when  and  only  when  the  curve  y  =f(x)  is  a  straight  line  ;  i.e. 


INFINITESIMALS   AND    DIFFERENTIALS  93 

when  f(x)  is  a  linear  function, 

f(x)  =  ax  +  b. 

Hitherto  x  has  been  taken  as  the  independent  variable,  Ax 
as  the  principal  infinitesimal.  We  come  now  to  the  theorem 
on  which  the  whole  value  of  differentials  for  the  purpose  of 
performing  differentiation  depends. 

Theorem.      The  relation  (4)  : 

dy  =  Dlydx, 

is  true,  even  when  x  and  y  ore  both  dependent  on  a  third  vari- 
able, t. 

Suppose,  namely,  that  x  and  y  come  to  us  as  functions  of  a 
third  variable,  t  : 

(6)  *  =  <K0>         y  =  iK9> 

and  that,  when  we  eliminate  t  between  these  two  equations,  we 
obtain  the  function  „.  v 

y  =/(»)• 

Then  dx  and  dy  have  the  following  values,  in  accordance  with 
the  above  definition,  since  t,  not  x,  is  now  the  independent 
variable,  A£  the  principal  infinitesimal : 

dy  =  Dty  At,  dx  =  Btx  M. 

We  wish  to  prove  that 

dy  =  Dxydx. 
Now  by  Theorem  V  of  Chap.  II,  §  5 : 

Dty  =  DxyDtx. 
Hence,  multiplying  through  by  At,  we  get : 
DtyM  =  Djj  ■  D.xAt, 
or  dy  =  Dxy  dx,  q.  e.  d. 


94  .     CALCULUS 

With  this  theorem  the  explicit  use  of  Theorem  V  in  Chap. 
II,  §  5  disappears,  Formula  V  of  that  theorem  now  taking  on 
the  form  of  an  algebraic  identity  : 

du  _  du  dy 
dx     dy  dx 

To  this  fact  is  due  the  chief  advantage  of  differentials  in  the 
technique  of  differentiation. 

Differentials  of  Higher  Order.  It  is  possible  to  introduce 
differentials  of  higher  order  by  a  similar  definition : 

(7)  d2y  =  D2yAx2,         cPy  =  D3y  Ax3,         etc., 

x  being  the  independent  variable.     We  should  then  have  by  (3) 

(8)  d2y=D2ydx2        or         ^=DJy,        etc. 

Unfortunately,  however,  relation  (8)  does  not  continue  to 

hold  when  x  and  y  both  depend  on  a  third  variable,  t.     For 

example,  suppose  , 

rr  x  =  t2y         y  =  a  +  t1. 

Then  y  =  a  +  x. 

When  t  is  taken  as  the  independent  variable,  we  have  ac- 
cording to  relation  (8) : 

d?y  =  D2ydt2  =  2dt2', 
and  since  dx  =  2tdt, 

it  follows  that  -         2dt2        1        1 


dx-     4t2dt-     2i2     2x 

On  the  other  hand,  when  x  is  taken  as  the  independent  vari- 
able, relation  (8)  becomes 

d?y  =  D2y  dx2  =  0, 
and  consequently  f,2 

dx2 


INFINITESIMALS   AND    DIFFERENTIALS  95 

Thus  the  quotient,  — ^,  is  seen  to  have  two  entirely  distinct 

values  according  as  t  or  x  is  taken  as  the  independent  variable. 
We  will  agree,  therefore,  to  discard  this  definition.  The  nota- 
tion — ^  as  meaning  Dx2y  is,  however,  universally  used  in  the 
Calculus,  and  so  we  will  accept  the  definitions 

&y  =  Dx%         ^=Dx*y,        etc., 
dx2         xy  dx*         *Si 

interpreting  the  left-hand  sides  of  these  equations,  however, 
not  as  ratios,  but  as  a  single,  homogeneous  (and  altogether 
clumsy !)  notation  for  that  which  is  expressed  more  simply  by 
Cauchy's  D. 

Remark.     The  operator  Dx  shall  be  written  when  desired  as 

—  •     Thus 
dx 


Dz—  appears  as 

D;-y  =  Dx{DTy) 


d      x 


a  —  x  dxa  —  x 

Again,  the  equation 
appears  as 


dhj  _  d  dy 
dx2     dx  dx 

Finally,  the  following  notation  is  sometimes  used : 

*3L  -  Dx2y  dx,         ^  =  Dx3y  dx,        etc. 
dx  dx2 

4.   Technique   of    Differentiation.     Consider,    for    example, 
Formula  II,  Chapter  II,  §  6 : 

Ds(u  +  v)  =  Dxu  +  Dxv. 

On  writing  this  formula  in  terms  of  differentials,  we  have 

d(u  -+-  v)  _du      dv 
dx  dx     dx 


96  CALCULUS 

Now  multiply  this  equation  through  by  dx : 

d(u  +  v)  =  du  +  dv. 

Hence  the  theorem:   The  differential  of  the  sum  of  two  functions 
is  equal  to  the  sum  of  the  differentials  of  these  functions. 

The  others  of  the  General  Formulas,  Chapter  II,  §§  6,  7, 
can  be  treated  in  a  similar  way  and  lead  to  corresponding 
theorems  in  differentials,  embodied  in  the  following  important 
group  of  formulas. 

General  Formulas  of  Differentiation-. 

I.  d(cu)  —  cdu. 

II.  d(u  +  v)  =  du  +  dv. 

III.  d(uv)  =  u  dv  +  v  du. 

TTT                                ,  u     vdu  —  udv 
L\.  d  —  = 

V  V1 

As  already  explained,  Theorem  V  reduces  to  an  obvious 
algebraic  identity  : 

dx      dy  dx7 

and  so  does  not  need  to  be  tabulated. 

Of  the  special  formulas  hitherto  considered,  only  two  need 
be  tabulated,  namely  : 

Special  Formulas  of  Differentiation. 

1.  dc  =  0. 

2.  dxn  =  nxn~1dx. 

The  first  of  these  formulas  says  that  the  differential  of  a 
constant  is  zero.  The  second  is  valid,  not  only  when  x  is  the 
independent  variable,  but  when  x  is  any  function  whatever  of 
the  independent  variable,  t.     Thus  if 

(1)  M=Vl^l 


INFINITESIMALS   AND    DIFFERENTIALS  97 

and  we  set 

(2)  x  =  l-t, 
equation  (1)  becomes 

(3)  u  =  xK 
Hence                                du  =  \  x~*  dx. 

But  dx  =  dl  +  d(—  t)=0  —  dt, 

and  thus                  _  dt  du  1 

du  = or 


2V1  -  t  dt         2 VI  -  t 

The  student  should  copy  off  neatly  on  a  card  the  size  of  a 
postal  the  General  Formulas  I-IV,  the  Special  Formulas  1., 
2.,  leaving  room  for  a  few  further  special  formulas.  All  the 
differentiations  of  the  elementary  function  of  the  Calculus  are 
based  on  these  two  groups  of  formulas. 

To  differentiate  a  function  means  henceforth  to  find  either 
its  derivative  or  its  differential.  Of  course,  when  one  of  these 
is  known,  the  other  can  be  found  by  merely  multiplying  or 
dividing  by  the  differential  of  the  independent  variable. 

We  proceed  to  show  by  a  few  typical  examples  how  differen- 
tials are  used  in  differentiation. 

Example  1.     Let  u  =  12  —  5x-\-7x*. 

To  find  du. 

Take  the  differential  of  each  side  of  this  equation,  and  apply 
at  the  same  time  Formula  II : 

du  =  d(12)+  d(-  5x)+  d(7x*). 

By  Formula  1,  d(12)  =  0. 

By  Formula  I, 

d(-5x)  =  -5dx      and       d(7xz)=7dx*. 

Hence  du  =  —  5  dx  +  21  x2  dx 

=  (-5  +  21x*)dx 

and  —  =  -5  +  21«2. 

dx 


98  CALCULUS 

These  steps  correspond  precisely  to  the  steps  the   student 

would  take  if  he  were  using  derivatives,  only  he  would  not 

have  written  thern  all  out  in  detail.     He  would  have  written 

down  at  sight :         _.  „      _ .    „ 

8  Dxu  =  —  o  +  21./-. 

He  can  avail  himself  of  the  facility  he  has  already  acquired 
and  shorten  the  work  as  follows.     Since 

du  =  Dxudx, 

he  can  begin  by  writing 

du  =(  )dx, 

and  then  fill  in  the  parenthesis  with  the  derivative.* 

Example  2.     Let  _  a2  —  x2 

a2  +  x2 
To  find  du. 

By  Formula  IV  we  have  :  • 

du  _  (tf-  4-  x*)rt(a?  -  x2)  -  (a2  -  x*)d(a?  +  J*) 
(a2  +  x2)2 

_  (a-  +  «;-)(—  2 xdx)  —  (a-  —  x-)(2xdx) 
(a2  +  x2)2 

4  a ".rtJ.r 


(a-  +  x)] 
du  4  aV 


r/.r  (a2  +  a?2)2 

The  student  would  probably  prefer  to  work  this  example  as 
follows.     Remembering  that 

du  =  D.u  dx, 

*  The  student  must  be  careful  not  to  omit  any  differentials.  If  one 
term  of  an  equation  has  a  differential  as  a  factor,  every  term  must  have 
a  differential  as  a  factor.     Such  an  equation  as 

du  =  —  6  +  21  x2 

is  absurd,  since  the  left-hand  side  is  an  infinitesimal  and  the  right-hand 
not.     Moreover,  there  is  no  such  thing  as  dxu. 


INFINITESIMALS   AND    DIFFERENTIALS  99 

begin  by  writing 

du  = dx, 

and  then  fill  in  the  fraction  by  the  old  familiar  methods  of 
Chapter  II. 

In  the  two  examples  just  considered,  the  processes  with 
differentials  correspond  precisely  to  those  with  derivatives 
with  which  the  student  is  already  familiar.  This  will  always 
be  true  in  any  differentiation  in  which  composite  functions  are 
not  involved ;  i.e.  whenever,  according  to  our  earlier  methods, 
the  vanished  Theorem  V  of  Chapter  II,  §  8  was  not  used.  It 
is  in  the  differentiation  of  composite  functions  that  the  method 
of  differentials  presents  advantages  over  the  earlier  method. 
We  turn  in  the  next  paragraphs  to  such  examples. 

EXERCISES 

Differentiate  each  of  the  following  functions  by  the  method 
of  differentials,  and  test  the  result  by  the  methods  of  Chap- 
ter II. 


1. 

u  =  x3  —  Sx  +  1. 

Ans.   du  =  3  xn-dx  —  3  dx. 

2. 

y  =  a  +  bx  +  ex-. 

Ans.   dy  =  b  dx  +  2  ex  dx. 

3. 

xo  =  a?  —  z3. 

Ans.    div  =  —  3  zHz. 

4. 

s  =  96t-16t2. 

Ans.    —=96-32*. 
dt 

5. 

s  =  v0t  +  ±gt2. 

Ans.    —  =  vQ  +  at. 
dt              y 

6. 

1-x 

u  = • 

A         -,         —2dx 
Ans.    du  =  —- 

i  +  x  (i  +  xy 

x  a         7       dx  —  x-  dx 

Ans.    dy  = 


J      1+a2  (1+a2)2 

1 4-  x  +  x2  A         7        x2  —  1  , 

8.  z  =    ^     ^     •  Ans.   dz  =  ———  dx. 

2x  2x* 

3  -  2  x  +  a;3  , .  a4  -  x* 

9.  u  = ■ — ! •  10.    y  = 


4  +  x-  -  x3  '       a4  +  a'x2  +  x* 


100  CALCULUS 

5.   Continuation.     Differentiation  of  Composite  Functions. 


Example  3.     Let       u  =  Vl  +  x  +  x2. 
To  find  —  • 

</.r 

Here,  we  begin  by  computing  du.     To  do  this,  introduce  a 
new  variable,  y,  setting 

y  =  1  +  x  ■+-  x2. 

Then  u  =  y*. 

Next,  take  the  differential  of  each  side  of  this  equation.     By 
Special  Formula  2  above, 


Moreover, 
Hence 

and 


Let  the  student  carry  through  the  above  differentiation  by 
the  methods  of  Chapter  II  and  compare  his  work  step  by  step 
with  the  foregoing.  He  will  find  that,  although  the  two 
methods  are  in  substance  the  same,  the  method  of  differentials 
is  simpler  in  form,  since  no  explicit  use  of  Theorem  V  here  is 
made. 

Abbreviated  Method  *  The  solution  by  differentials  can  be 
still  further  abbreviated  by  not  introducing  explicitly  a  new 

*  The  student  should  not  hasten  to  take  this  step  himself.  He  will  do 
well  to  omit  the  text  that  follows  till  he  has  worked  a  score  or  more  of 
problems  in  differentiating  composite  functions  as  set  forth  under  Ex- 
ample 3,  introducing  each  time  explicitly  a  new  variable,  as  y.  z,  etc. 
Not  until  he  comes  himself  to  feel  that  the  abbreviation  is  an  aid,  should 
he  attempt  to  use  it. 


du  = 

:dy?=\ 

y  Uly. 

dy. 

=  (1+2 

x)dx. 

du  — 

.    (1  +  2 

x)dx 

2V1  + 

X  +  .'' 

du 

l  +  2a 

dx 

2V1  + 

X  +  X2 

INFINITESIMALS   AND    DIFFERENTIALS  '.   f   101 

variable,  y.     The  problem  is  to  find  du,  wbjen    .< 

u  =(1  +  x  +  x2)\ 

Now,  Special  Formula  2,  as  has  already  been  pointed  out, 
holds,  not  merely  when  x  is  the  independent  variable,  but  for 
any  function  whatsoever.  It  might,  for  example,  equally  well 
be  written  in  the  form  : 

d  [>(»)]"  =  ji[</>(»]"-i  d<l>(x). 

In  the  present  case,  then,  the  content  of  that  theorem,  —  the 
essential  and  complete  truth  it  contains,  —  enables  us  to  write 
down  at  once  the  equation  : 

d(l  +  x  +  xrf  =  |(1  +  x  +-  x^  d(l  +  x  +  a?2). 

This  last  differential  is  computed  at  sight,  and  thus  the 
answer  is  obtained  in  two  steps. 

Even  these  two  steps  are  carried  out  mentally  as  a  single 
process,  when  the  student  has  reached  the  highest  point  in  the 
technique  of  differentiation.     He  then  thinks  of  the  formula : 

j  r        dx 

2Vx 

realizes  that  it  holds,  not  merely  when  x  is  the  independent 
variable,  but  for  any  function  of  x,  and  so  writes  down  first 
the  easy  part  of  the  right-hand  side  of  the  equation,  thus : 


d  Vl  +  x  +  a?  = 


2  Vl  +  x  +  x2 


carrying  in  his  head  the  fact  that  the  numerator  is  the  differen- 
tial of  the  radicand,  i.e.  d(l  +  x  +  a;2).  This  differentiation  he 
performs  mentally,  and  thus  has  the  final  answer  with  no 
intermediate  work  on  paper  : 

dVl^.4-^^     (l+g»)**    . 

2Vl  +  x  +  a2 


102  CALCULUS 

Example  4..  The  method  of  differentials  is  especially  use- 
ful in  the  case  of  implicit  functions.  Thus,  to  find  the  deriva- 
tive of  y  with  respect  to  x  when 

x?-3xy  +  2yi  =  l. 

Take  the  differential  of  each  side  : 

3  x2  dx  —  3  x  dy  —  3  y  dx  +  8  y3  dy  =  0. 

Next,  collect  the  terms  in  dx  by  themselves  ;    the  others  will 
contain  dy  as  a  factor : 

(3x2  -  3y)dx  +  (Hif  -  3x)dy  =  0. 

Hence  <%  =  3y-3a* 

dx     8^  —  3oj 

EXERCISES 

Differentiate  the  following  twelve  functions  by  the  method 
of  differentials  and  also  by  the  methods  of  Chapter  II  (in 
either  order),  introducing  each  time  explicit!)/  the  auxiliary 
variable,  if  one  is  used. 

1.  w=Va4  +  a2a;2  +  x4.  Ans.   du  =    (a2x  +  2x")dx   . 

Va4  +  a2a-2  +  x* 

2.  y  =  —  -  •  Ans.   dy  = 


Vl-*2  (l-a*)* 

o  1  A  T  dX 

3.    ?<  = •  Ans.    du  = 


1  -  x  (1  -  xy 

Suggestion.      Introduce    an    auxiliary    variable    y  =  1  —  x. 
Then  ?<  =  y_1. 

1  .         du  2 

4.   u  =  — •  ^4?is.    — 


(1  -  xy  dx      (1  -  a;)8 

_                1  .         dv        —  2<c 

5.   y  =  - Ans.    s.  = 

1  +  a:2  dx      (1  +  x2)2 


INFINITESIMALS   AND    DIFFERENTIALS         103 

6.  s=      a2      ■  Ans.    (l?  = *£-. 

(a  + 1)2  dt  (a  +  03 

7.  2x2-xy  +  ly2  =  o.  Ans.    ^  =  4a;~y. 

clx     x  —  Sy 

8.  xy  =  a2.  Ans.    -J  —  —  -f . 

(I.r  x 

9.  y2  =  2mx.,  Ans.    (_K=— . 

dx       y 

10. h  —  =  1-  -l"s-    —  = 

a2      62  efcc  r  >'-'// 

11.  2z2  +  3v2  =  10.  ^4»s.    ^  =  -^. 

12.  2  aw  —  x  +  y  =  0.  Ans.    ~¥  = 


dx     2  x  -f  -  1 

The  student  can  work  the  problems  at  the  end  of  Chapter 
II  by  the  method  of  differentials.  For  further  practice,  if  de- 
sired, the  following  examples  are  appended. 

du      7xi-2x       I 


13.  u  =  (x2  +  1)  V.r3  —  x.  Ans.    —  = 

dx         2v'.r;!  -  .'• 

14.  y=(x  +  2b)(x-bf.  Ans.    ^  =  3  (x*<  -  b2). 

15.  «= Ans. 


Va2  —  a2  rfa;      V(a2  -  a;)3 

la  —  x  a        du                  a 

16.  ?t  =  \/ Ans.    —  = 

v     x  dx         2xVax  —  x- 

x  —  a  A        du 

17.  u= —  .                             Ans.    —  =  ■ 


&\3 


^2  ax  —  x2  dx      V(2ax  —  x2) 

18-    n  =  (~       -]  AnS-    dx  =  2x^? 


X 


A/4  _i_  i,i\2  dz     Ays—'^ 

19.    z=  y  ^      )  ■  Ans.    —  =4^— .     • 

f-    J  dy  y* 


104 


20.   „-?*±*Vtf=rf. 

x3 


lx2 


V  +  a  +  1 

22.  tt= (ft -*»)*. 


CALC 

IULUS 

du  _ 
dx 

Ans. 

du 
dx 

3a< 

i 

-X2. 

A)lS. 

x4  Va'- 
3  a4 

-a;2 

(x~+ 
Ans. 

x  + 

du 
dx 

l)V<e*  +  a? 

+  1 

8Vx  — 
3  a,-4 

x3 

/  ;        Im                                   a        du     4(jb*—  a3)3 
23.    w=(a;*—  a*)4.  -4«s.    —  =  -* — L 


dx 


3x* 


24 .    x  =  x  (x3  +   m  ' .  ^Ins.    *? :  =  5  (a3  +  1)  (x3  +  5)i 

J  dx  N  x 


CHAPTER   V 
TRIGONOMETRIC   FUNCTIONS 

1.  Radian  Measure.  In  Trigonometry,  the  radian  measure 
of  an  angle  was  introduced,  apparently  for  no  good  purpose. 
The  reason  lies  in  the  importance  for  the  Calculus  of  this  new 
system  of  measurement,  and  will  become  clear  in  the  next 
paragraph,  when  we  come  to  differentiate  the  sine.  We  will 
first  recall  the  definition. 

Let  a  circle  be  described  with  its  centre  at  the  vertex  0  of 
the  angle ;  let  r  denote  the  length  of  the  radius  of  the  circle 
and  s,  that  of  the  intercepted  arc.     Then  \ 

the  radian  measure,  $,  of  the  angle  is 
defined  as  the  ratio  s/r : 


(1) 


Fig.  34 


For  a  right  angle,  s  =  ~,  and  hence  6  =  ^-     A  straight  angle 
has  the  measure 

6  =  7T  =  3.14159  26535  89793  ■  •  -. 


Let  cji  be  the  measure  of  the  given  angle  in  degrees.     Then 
0  and  <f>  are  proportional, 

6  =  ccf>, 

where*  c  is  a  constant.  To  determine  c,  use  a  convenient  angle 
whose  measure  is  known  in  both  systems ;  for  example,  a 
straight  angle.     For  the  latter, 


and 
105 


<f>  =  180. 


106  CALCULUS 

Substituting  these  values  in  the  above  equation  we  find : 

7r=c180'      c=iio' 

and  hence 

(2)  .-^         *=A 

This  equation  can  also  be  written  in  the  form 

(3)  l=*l 

w  T      180- 

and  thus  an  easily  remembered  rule  of  conversion  from  radian 

measure  to  degree   measure,  or  the  opposite,  obtained :    TV 

radian  measure  of  an  angle  is  to  -n-  as  its  degree  measure  is  to  180. 

The  unit  of  angle  in  radian  measure,  i.e.  the  angle  for  which 

0  —  1         and  hence         s  =  /•, 

is  called  the  radian.  It  is  obvious  geometrically  that  it  is  a 
little  less  than  60°.  Its  precise  value  (to  hundredths  of  a  sec- 
ond) is  given  by  (2)  : 

<£!e=1  =  1S0  =  r.7°  IT'  44.81"  (=57.29578°). 

7T 

On  the  other  hand,  the  radian  measure  of  an  angle  of  1°  is 

6  L_,  =  -2L  =  .01745     32925     19943  .... 
I  o-i      lg0 

The  student  should  practice  impressing  the  more  important 
angles,  as  30°,  45°,  60°,  90°,  120°,  etc.,  in  radian  measure  until 
he  is  thoroughly  familiar  with  the  new  representation  for 
them. 

If,  in  particular,  the  radius  of  the  circle  is  taken  as  unity, 
then  6  and  s  are  the  same  number  : 

(4)  6  =  s,  when  r  =  1  ; 

or  the  arc  is  equal  to  the  angle.  Thus  the  radian  measure  of  an 
angle  might  have  been  defined  as  the  length  of  the  intercepted 


TRIGONOMETRIC   FUNCTIONS  107 

arc  in  the  unit  circle  {i.e.  the  circle  of  unit  radius  with  its 
centre  at  0). 

Graph  of  sin  x.     It  is  important  for  the  student  to  make  an 
accurately  drawn  graph  of  the  function 

y  =  sin  x, 

x  being  taken  in  radian  measure.  Let  the  unit  of  length,  as 
usual,  be  the  same  on  both  axes,  and  let  it  be  chosen  as  1  cm. 
For  this  purpose  Peirce's  Table  of  Integrals  (the  table  of 
Trigonometric  Functions  near  the  end)  is  especially  convenient, 
since  the  outside  column  gives  the  angles  in  radian  measure, 
and  thus  as  many  points  of  the  graph  as  are  desired  can  be 
plotted  directly  from  the  tables. 

j/=sm  x 


Fig.  35 

Since  sin  (tt  —  x)  =  sin  x 

each  determination  of  the  coordinates  (x,  y)  of  a  point  on  the 

graph,  for   which   0  <  x  <  -  yields  at  once  a  second   point, 

namely  (tt  —  x,  y).  Thus  one  arch  of  the  curve  is  readily  con- 
structed from  the  Tables.* 

From  this  arch  a  templet,  or  curved  ruler,  is  made  as  fol- 
lows. Lay  a  card  under  the  arch  and  with  a  needle  prick 
through  enough  points  so  that  the  templet  can  be  cut  ac- 
i-i irately  with  the  scissors. 

By  means  of  the  templet  further  arches  can  be  drawn  me- 
chanically, and  thus  the  curve  is  readily  continued    in    both 

*  The  graph  could  be  made  directly  without  tables  from  purely  geomet- 
rical considerations.  Draw  a  circle  of  unit  radius.  Construct  geomet- 
rically convenient  angles,  as  those  obtained  from  a  right  angle  by 
successive  bisectors.  Measure  any  one  of  these  angles,  "%.ABP„,  in  ra- 
dians and  this  number  will  be  the  abscissa  of  the  point  on  the  graph,  the 


108 


CALCULUS 


directions  to  the  edges  of  the  paper.*     Put  this  curve  in  the 

upper  quarter  of  a  sheet  of  centimetre  paper. 

The  graph  brings  out  clearly  the  property  of  the  function 

expressed   by  the  word  periodic.     The    function   admits   the 

»( Hod  2ir,  since  .     .        „    . 

sm  (x  +  2  7t)  =  sin  x 

Graph  of  cos  a;.     By  means  of  the  templet  the  graph  of  the 

function 

y  —  COS  X 

can  now  be  drawn  mechanically.     This  function  also  admits 

the  period  2  n  : 

cos  (x  +  2  it)  =  cos  x. 

y  =  COSX 


ordinate  being  the  perpendicular  dropped  from  P„  on  the  line  BA. 
if  n  =  3,  the  coordinates  of  the  point  on  the  graph  are  : 


Thus, 


Fig.  36 


x  =  — =  1.18,       ?/ =  .92. 
8 

A  second  point  of  the  arch. 
that  corresponding  to  P5,  has 
the  same  y,  its  coordinate  be- 
ing 

X  =  ir-  —  =1.96,         ?/=.92. 


Of  course,  the  distance  ir  must  be  laid  off  on  the  axis  of  x  by  measure- 
ment :  ii  cannot  be  constructed  geometrically  from  the  unit  length.  This 
done,  the  further  abscissae  are  found  by  successive  bisectors. 

•  In  order  to  obtain  the  most  satisfactory  figure,  observe  that  the 
curve  has  a  point  of  inflection  at  each  of  its  intersections  with  the  axis  of 
X,  the  tangent  there  making  an  angle  of  ±  45°  with  that  axis.  Since  a 
curve  separates  very  slowly  from  an  inflectional  tangent,  it  will  be  well  to 
draw  these  tangents  with  a  ruler.  <  >n  laying  down  the  templet,  the  curve 
can  then  be  ruled  in  from  the  latter  with  great  accuracy.  It  will  not 
sensibly  from  its  tangent  for  a  considerable  distance  from  a 
point  of  inflection. 


TRIGONOMETRIC   FUNCTIONS 


109 


Put  the  graph  in  the  second  quarter  of  the  sheet,  choosing  the 
axis  of  y  for  this  curve  in  the  same  vertical  line  as  the  axis  of 
y  for  the  sine  curve  above.  There  remains  the  lower  half  of 
the  sheet  for  the  next  graph. 

Graph  of  tan  x.  The  same  tables  make  it  easy  to  plot  points 
profusely  on  the  graph  of  the  function 

y  =  tan  x 
Take  the  axis  of  y  in  the  same  ver- 


in  the  interval  0<  x  < 


—  9 


j/=tani 


Fig.  38 


tical  line  as  in  the  case  of  the  preceding  graphs.     This  done,  a 
second  templet  is  made  and  by  means  of  it  the  graph  is  drawn 

mechanically  for  values  of  x  such  that  —  ^<  x  <  0. 

It  is  desirable  furthermore  to  plot  the  function  in  the  two 
adjacent  intervals 


110 


CALCULUS 


2V  2  ' 


3  7T     ^  .  7T 

-T<x<-2' 


in   order   to   suggest   the   fact  that  this  function  admits  the 

period  it  :  .  . 

tan  (.r  -f  w)  =  tan  x. 

2.    Differentiation  of  si  n  x.     To  differentiate  the  function 

(1)  ?/  =  sin.r, 

apply  the  definition  of  a  derivative  given  in  Chap.   II,  §  1. 

Give  to  ./•  an  arbitrary  value 
.>•„  ami  compute  the  corre- 
sponding value  y0  of  y  ; 

y0  =  sin  x0. 

Then  give  x  an  increment  A.r, 
and  compute  again  the  corre- 
sponding value  of  y : 

Uo  +  Ay  =  sin  (cc0  4-  Aa). 
Hence 

A//  =  sin  (.r0  +  Ax)  —  sin  x0, 


31'  31 


Fig.  30 


(2) 


Ay  _  sin  (.r0  +  Ax)  —  sin  x0 
Ax  Ax 


It  is  at  this  point  in  the  process  that  the  specific  properties 
of  the  function  sin  x  come  into  play.  Here,  the  representa- 
tion of  sin  x  by  means  of  the  unit  circle,  familiar  from  the 
beginning  of  Trigonometry,  is  the  key  to  the  solution.  From 
the  figure  it  is  clear  that 

sin  x0  =  MP,  sin  (x0  +  A.r)  =  M'P', 

Ay  =  sin  (x0  +  Ax)  —  sin  x0  =  QP',  Ax  =  PP\ 


Hence 
(3) 


A1=QP' 
Ax     pp> 


TRIGONOMETRIC   FUNCTIONS  111 

and  so  we  want  to  know  the  limit  approached  by  the  latter 
ratio:  _ 

P'=P  ppi 

By  virtue  of  the  Fundamental  Theorem  of  Chap.  IV,  §  2,  we 
can  replace  this  ratio  by  a  simpler  one,  since  the  arc  PP  and 
the  chord  PP'  are  equivalent  infinitesimals  :  * 

P~P' 

lim =  1. 

PP' 

Hence  lim  -^-^  =  lim  " — • 

p'=p  ppi      p'=p  ppi 

On  the  other  hand,  the  triangle  QPP'  is  a  triangle  of  refer- 
ence for  the  %.  QPP'  =  <f>,  and  so 

QP'      ■   j. 

-g —  =  sin  <£. 
PP 

When  P'  approaches  P,  the  secant  PP'  (i.e.  the  indefinite  liue 
determined  by  the  two  points  P  and  P')  approaches  the  tan- 
gent PT  at  P,  and  thus 

P'=P  L 

Finally,  then, 

lim  ^^  =  lim  sin  </>  =  sinf  - 
P'=p  ppi     p±p  V  2 

and  consequently  t     Ay 

^  J  lim  — ^  =  cos  x0, 

&x±o  Ax 

*  The  student  should  assure  himself  of  the  truth  of  this  statement  by 
visualizing  the  figure  (making  an  accurate  drawing  with  ruler  and  com- 
pass for  angles  of  30°,  15°,  and  7|°,  the  circle  used  being  10  in.  in 
diameter)  and  realizing  that,  when  P'  is  near  P,  the  difference  in  length 
between  the  arc  and  the  chord  is  but  a  minute  per  cent  of  the  length  of 
either  one.     A  formal  proof  will  be  found  below. 


Xn    =  COS  Xn 


112  CALCULUS 

or,  on  dropping  the  subscript, 

(4)  Z)xsin  x  =  cos  x. 

This  theorem  gives  rise  to  the  following  theorem  in  dif- 
ferentials : 

(5)  dsin  x  =  cos  xdx. 

Reason  for  the  Radian.  The  reason  for  measuring  angles  in 
terms  of  the  radian  as  the  unit  now  becomes  clear.  Had  we 
used  the  degree,  the  increment  A.e  would  not  have  been  equal 
to  PP' ;  we  should  have  had  : 

Ax      PP'                  .         180  £&, 
=      — ,        or        Ax  = PP. 

360       2tt  it 

Hence  (3)  would  have  read  : 

Ay=   it     QP 

A./:       180  '  pp? 
and  thus  the  formula  of  differentiation  would  have  become: 

D,  sin  x  =  -^—  cos  x. 
180 

The  saving  of  labor  in  not  being  obliged  to  multiply  by  this 
constant  each  time  we  differentiate  is  great.  Still  more  impor- 
tant, however,  is  the  elimination  of  a  multiplier  which  is  of 
the  nature  of  an  extraneous  constant,  whose  presence  would 
have  obscured  the  essential  simplicity  of  the  formulas  of  the 
Calculus. 

EXERCISE 

Prove  in  a  similar  manner  that 

D„  cos  x  =  —  sin  x. 

3.  Certain  Limits.  In  the  foregoing  paragraph  we  have  made 
use  of  the  fact  that  the  ratio  of  the  arc  to  the  chord  approaches 
1  as  its  limit.     A  formal  proof  of  this  theorem,  based  on  the 


TRIGONOMETRIC   FUNCTIONS 


113 


axioms  of  geometry,  can  be  given  as  follows.  Draw  the  tan- 
gent at  P  and  erect  a  perpendicular  at  P  cutting  the  tangent 
in   Q.     Denote  the  angle  ^  PPQ  by  a. 

Then       . w 

PP  <PP  <PQ  +  PQ; 


for  i)  a  straight  line  is  the  shortest  dis- 
tance between  two  points  ;  and  ii)  a  convex  curved  line  is  less 
than  a  convex  broken  line  which  envelops  it  and  has  the  same 
extremities.     But 


PQ  = 


PP' 

cos  a 


P  Q  =  PP  tan  a. 


Hence 


PP' 


pp<       cos  a 


+  tan  a. 


When  a  approaches  0,  the  right-hand  member  of  the  double 
inequality  approaches  1 ;  hence  the  middle  member  must  also 
approach  1,  or  w 

lim =  1, 

PP 


q.  e.  d. 


The  foregoing  proof  holds,  not  merely  for  a  circle,  but  for 
any  curve  with  a  convex  arc  PP.  Consequently  the  theorem 
is  established  generally. 


TJie  Limit  lim 


since 


it  is  clear  that 

MP=  sin  «, 

and  hence 


From  Fig.  41 

AP=a, 

sin  a  _  MP 
a        ^P 


Fig.  41 


MA 


:i) 


By  direct  inspection  of  the  figure  it  is  seen,  then,  that 
sin  a 


lim 


=  1. 


114 


CALCULUS 


A  formal  proof  of  this  equation  can  be  given  as  follows. 
From  Fig.  42 

PP'  =  2  sin  a,         PP'  =  2a. 


Hence 


Hin  a  =  PP' 
a        pp' 


and  therefore,  by  the  proposition  just  established, 

PP' 


,.     sin « 

lim =  hm 


=  1. 


>PP/ 


Another  Proof  of  (1).     The  area  of  the  sector  OAP,  Fig.  41', 
is  i«,  and  it  obviously  lies  between  the  areas  of  the  triangles 
OMP  and  OPX.     Hence 
\p' 

\  sin  «  cos  «  <  \  a  <  V  tan  <c 

Dr  cos  «  <  — < 


Fir,.  42 


sin  «      cos  a 

fr»-       When  «  approaches  0,  each  of  the  ex- 
treme terms  approaches  1,  and  so  the 
middle  term  must  also  do  so,  q.  e.  d. 
From  Peirce's  Tables,  p.  130,  we  see  that 

sin  4°  40'  =  .0814, 

and  the  same  angle,  measured  in  radians,  also  has  the  value 
.0814,  to  three  significant  figures.  Thus  for  values  of  «  not 
exceeding  .0814,  sin  «  differs  from  a  by  less  than  one  part  in 
800,  or  one-eighth  of  one  per  cent. 

TJie  Limits  lim  1  ~  cos  a  and  lim1"008"-     From  Fig.  42, 
a±o  a  °=o  «2 

the  first  of  these  limits  is  seen  to  have  the  value  0 : 


, .      1  —  cos  a      n 
lim =  0. 


TRIGONOMETRIC   FUNCTIONS 


115 


A  formal  proof  can  be  derived  at  once  by  the  method  em- 
ployed in  the  evaluation  of  the  next  limit, 

1  —  cos  a 


lim 


Expressing  1  —  cos  a  in  terms  of  the  half  angle,  we  have 
1  —  cosoc  =  2  sin2r- 


Hence 


and 


1  —  cos , 


>  « 
_2_1 

'      ~2 


•    a 

sin- 

2 

a 
2 

lim 

oil) 


1—  cosa_l  y 
a2  2  a^=u 


a 

2 


EXERCISES 


In  the  accompanying  figure  determine  the  following  limits 
when  a  approaches  0  : 


1.    lim 


AR 


MP 

2.    lim^-- 
AP 

RQ 


Ans.    - . 

2 


Ans.    1. 


k 

Q 

1^ 

\S 

M 

V 

o*^ 

dl 

FIg.  43 


3.    lim 


6.    lim 


MP 

MA 
PQ 


4.    lim 


7.    lim 


RP_ 
AP 

PQ 

AN' 


5.    lim 


8.    lim 


PN 
AP 

RQ 
PN 


Determine  the  principal  part  of  each  of  the  following  infini- 
tesimals, referred  to  a  as  principal  infinitesimal : 

9.    MP.    Ans.  a.        10.  PR.  Ans.^a.    11.    RQ. 
12.    PN.  13.  AQ.  14.    MA.    Ans.  i«l 

15.    PQ.  16.  MN  17.    AQ-MP. 


1 1 6  CALCULUS 

4.  Critique  of  the  Foregoing1  Differentiation.  The  differenti- 
ation of  sin  x  as  given  in  §  1  has  the  advantage  of  being  direct 
and  lucid,  and  thus  easily  remembered.  Each  analytic  step  is 
mirrored  in  a  simple  geometric  construction.  It  has  the  dis- 
advantage, however,  of  incompleteness.  For,  first,  we  have 
allowed  Ax,  in  approaching  0,  to  pass  only  through  positive 
values ;  and  secondly  we  have  assumed  x0  to  lie  between  0  and 
^  7r.     Hence  there  are  in  all  seven  more  cases  to  consider. 

An  analytic  method  that  is  simple  and  at  the  same  time 
general  is  the  following.  Recall  the  Addition  Theorem  for 
the  sine : 

sin  (a  +  b)  =  sin  a  cos  b  +  cos  a  sin  b, 

sin  (a  —  6)  =  sin  a  cos  b  —  cos  a  sin  b, 

whence  sin  (a  +  b)—  sin  (a  —  6)  =  2  cos  a  sin  b. 

Let  a  +  &  =  a?0  -f  Ax,  a  —  b  —  .<•„. 

Solving  these  last  equations  for  a  and  b,  we  get : 

a  =  .r„  -f  o  = 

2  2 

Thus      sin  (a0  +  A.r)  —  sin  x0  =  2  cos  (  x0  -J-—  J  sin  -— , 

and  the  difference-quotient  becomes 

•    Ax 

Sill 

A'/  /  Ax\         '1 

—  =  cos     .<•,,  4- 

Aa:  \°^  'J  J     Ax 

2 

The  first  factor  on  the  right  approaches  the  limit  cosa\) 
when  Ax  approaches  0.  On  setting  }rAx=a,  the  second  fac- 
tor becomes 

Sill  (C 

a 

Hence  the  factor  approaches  1.     Thus 

r      A?/ 
hm-^.=  cos#0, 

Ax=yi  A.c 


TRIGONOMETRIC   FUNCTIONS  117 

or,  on  dropping  the  subscript, 

Dx  sin  x  =  cos  x. 

5.    Differentiation  of  cos  x,  tan  x,  etc.     To  differentiate  the 
function  cos  x,  introduce  a  new  variable,  y,  by  the  equation 

y  =  —  —  x.       Hence       x  =  - —  y, 
and  cos  x  =  cos  (  -  —  y  )=  sin  y. 


Taking  the  differential  of  each  side  of  the  equation  thus  ob- 
tained, we  have : 

d  cos  x  =  d  sin  y  =  cos  y  dy. 

But  cos  y  =  sin  x         and         dy  =  —  (fcc. 

Hence 

(1)  d  cos  a;  =  —  sin  x  dx. 

To  differentiate  the  function  tan  x,  set 

sin  a: 
tan  x  = 


Hence  d  tan  a;  = 


cosa; 
cos  a-rfsin  x  —  sin  xd  cos  a? 

COS2  X 

cos  -  a;  dx  +  sin2  a*  da;        da: 


cos2  x  cos2  x 

and  thus 

(2)  d  tan  a;  =  sec2  a:  da;. 

It  is  shown  in  a  similar  manner  (or  by  setting  x  =  —  —  y  in 
the  equation  just  deduced)  that 

(3)  d  cot  x  =  —  esc2  x  dx. 

These  are  the  important  formulas  of  differentiation  for  the 
trigonometric  functions.  By  means  of  them  all  other  differen- 
tiations of  these  functions  can  be  readily  performed.     Thus, 


118  CALCULUS 

to  differentiate  the  function  sec  x,  set 

sec  x  =  (cos  x)~\ 

mi  ,  dcosx      sinxdx 

Then  aseca  =  —  —  =-       — . 

cos2  x        cos2  x 

It  is  not  desirable  to  tabulate  the  result,  since  one  rarely 
has  occasion  to  differentiate  either  sec  x  or  esc  x,  and  when  the 
occasion  does  arise,  the  differentiation  can  be  worked  out 
directly,  as  above. 

The  student  should  now  add  to  his  card  of  Special  Formulas 
the  four  main  formulas  just  obtained.  This  card  will  now  read 
as  follows : 

1.  dc  =  0. 

2.  dxn  =  nx71'1  dx. 
d  sin  x  =  cos  x  dx. 

4.  d  cos  x  =  —  sin  x  dx. 

5.  d  tan  x  =  sec2  x  dx. 
G.  d  cot  x  =  —  esc2  x  dx. 

6.  Shop  Work.  To  acquire  facility  in  the  use  of  the  new 
results,  the  student  should  work  a  generous  number  of  simple 
examples,  for  which  the  following  are  typical. 


Example 

1.     Tod 

lifferentiate  the  function 

u  =  sin  ax. 

Let 

y  =  ax. 

Then 

u  =  sin  y, 

and 

du  =  d  sin  y  =  cos  y  dy. 

But 

dy  =  a  dx. 

Hence, 

substituting, 

we  have 

du 

,                     a    . 
=  a  cos  axdx        or         —  sin  ax  =  a  eos  ax. 
dx 

TRIGONOMETRIC   FUNCTIONS  119 

The  solution  can  be  abbreviated  as  follows.     The  equation 

d  sin  x  =  cos  x  dx 

is  true,  not  merely  when  x  is  the  independent  variable.     It 
holds,  for  example,  in  the  form 

d  sin  y  =  cos  y  dy, 

where  y  is  any  function  of  x.     Hence  we  can  write  imn^ediately 

dsin  ax  =  cos  ax  d(ax), 

and  thus  obtain  the  result 

d  sin  ax  =  a  cos  ax  dx. 

Example  2.     To  differentiate  the  function 
u  =  Vl  —  k2  sin2  cf>. 
Let  z  =  1  —  k2  sin2  <£. 

Then  u  =  z* 

du  =dz*  =  ±z~~2dz; 
dz  =  —  k2d  sin2  <£. 
Let  y  =  sin  <f>. 

Then  dy  =  cos  <f>  d<f> 

and  d sin2  (f>  =  d y2  =  2y  dy  =  2  sin  <£  cos  <£ d<f>. 

Hence  du  =±z~^(—  2&2sin  <j>  cos  (f>d<f>) 

du  k2  sin  <f>  cos  <b 

—  = —J—. 

d$  Vl  -  k2  sin2  <£ 

Example  3.     If      sin  ^  +  sin  y  =  x  _  y> 

!  to  find  -^.     Take  the  differential  of  each  side  of  the  equation : 
dx 

cos  x  dx  +  cos  y  dy  =  dx  —  dy. 
Hence  (cos  x  —  l)dx  +  (cos  y  +  l)dy  =  0 

dy  _  1  —  cos  x 


and 


da;     1  +  cos  y 


120  CALCULUS 

EXERCISES 

Differentiate  the  following  functions. 

du 
1.    u  =  cos  ax. 


2. 

y  =  COS2  X. 

3. 

y  =  esc  x. 

4. 

X 

u  =  tan  — 
2 

5. 

u  =  cot  2x. 

6. 

?/  =  sec  3  x. 

8. 

u  =  sin3  x. 

10. 

u  =  x  +  tan  x. 

12. 

u  =  sec2  x. 

14. 

sin  # 

1  —  cos  X 

15. 

u  =  Vl  +  cos  X. 

16. 

1  —  COS  X 
u  = 

1  +  COS  X 

18. 

sinx 

a  +  &  cos  x 

20. 

1 

sin  x  +  cos  a; 

22. 

*   u  =  vers  #. 

dx 

—  a  sin  ax. 

dy  _ 
dx 

—  2  sin  x  cos  x- 

dx 

csc2  X  cos  X. 

du  _ 
dx 

1  9  X 

-  sec2 — 

2  2 

du 

dx 

-2  csc2  2  a;. 

7 

.    u  = 

tan2  ax. 

0 

.    »/  = 

;  1  —  sin  x. 

11 

.    u  = 

:  COS3  X. 

13 

.     >l  = 

sin  .V  cos  x. 

du         1       ,x 

—  = esc2  -  • 

dx         2        2 

du  1  x 

—  = sin-- 

dx          A/2        2 

,  _  1  +  sin  x 

17.     M  = 


19.    u  = 


21.    t«  = 


—  sinx 

1 


a  cos  x  4-  b  sin  x 
1 


(a-|-6  cos  x)2 

du 

—  =  sin  x. 
dx 

.  du 

23.*    ?<  =  covers  x.  —  =  —  cos  x. 

dx 

*  The  versed  sine  and  the  coversed  sine  are  defined  as  follows  : 
vers  x  —  1  —  cos  x  ;  covers  x  =  1  —  sin  x. 


TRIGONOMETRIC   FUNCTIONS  121 


x 

cos  - 

2 

24.    u  =  x  sin  2x.  25.    u 


x 


26.    u 


=  tan(l-l)  27.    «- «*(§-=)■ 


**r\  i.  "^  ^-krt.  Sill  irX 

28.    w=tan- 29.  u  = 

1  —  x  x 

30.    %i  —  sin  x  -\-  cos  2  x.  31.  m =  a;2  cds  nx. 

32.  1  33.  u  = 


VI  -  &2  sin'2  <f>  Vl  —  A;2  sin2  <£ 

n.  •    /     ,     ■>       dv  cos  (x  +  ?/)  —  cos  y 

34.  ^cosy=  sin  (a  +  y).      -?-  = ^ — ±-d± £_. 

dx  sin  (x  +  2/)  +  x  sin  y 

35.  tanx—  cot  y= sin  a;  sin?/.         36.    sin  x  +  sin  y  =  1. 
37.    tan  6  +  tan  <f>  =  2  tan  <£  tan  6.  38.    x  =  ysin?/. 

7.  Maxima  and  Minima.  By  means  of  the  new  functions 
studied  in  this  chapter  the  range  of  problems  in  maxima  and 
minima  which  can  be  treated  by  the  Calculus  has  been  materi- 
ally enlarged.  No  new  principles  are  involved ;  the  student 
should  go  over  carefully  the  paragraphs  of  Chap.  Ill  relating 
to  this  subject,  before  he  proceeds  farther  with  the  present 
paragraph. 

Example  1.  A  man  in  a  rowboat  1  mile  off  shore  wishes  to 
go  to  a  point  which  is  2  miles  inland  and  4  miles  up  the 
beach.  If  he  can  row  at  the  rate  of  5  miles  an  hour,  but  can 
walk  only  3  miles  an  hour  after  he  lands,  in  what  direction 
should  he  row  in  order  to  get  to  his  destination  in  the  shortest 
possible  time  ? 

In  the  first  place,  it  is  clear  that  the  straight  line  AEB  is 
not  the  best  path.  For,  if  he  rows  toward  a  point  P  slightly 
farther  up  the  beach,  the  amount  by  which  he  lengthens  the 
leg  AP  of  his  path  is  very  nearly  equal  to  the  amount  by  which 


122 


CALCULUS 


he  shortens  the  leg  PB*     Consequently  the  time  is  short- 
ened. 

On  the  other  hand,  P  obviously  ought  not  to  be  taken  so 

far  up  the  beach  as  D. 
The  minimum  occurs, 
therefore,  for  some  in- 
termediate point. 

Let  the  angles  8,  </> 

be  taken   as   indicated 

in    the    figure.     Then, 

s 


n 

/  / 

o          e/ 

t/ 

_> 

''    ^-^d 

p             i) 

1 

SJ&c   ° 

A 

since  t  — 


Fig.  44 


.IP 


1 


time  from  A  to  P  =  -  —  =  — 

5       5  cos  8 

time  from  P  to  B  = 


3        3  cos<^ 

Hence  the  total  time,  u,  which  is  to  be  made  a  minimum  is 

1  2 


(1) 


u  =  — h 

5cos0      3cos<£ 


Moreover,  8  and  <£  are  connected  with  each  other  by  a  rela- 
tion which  is  readily  obtained  by  expressing  the  distance  CD 
in  two  ways  : 

(2)  tan  8  +  2  tan  c£  =  4. 

We  are  now  ready  to  compute  du/dd  and  set  it  equal  to  0 : 

,    _  _  d  cos  8  _  2  d  cos  <f> 
5  cos-  8      3  cos2  <j> 


(3) 


sec2  6  sin  8  in  .  2  sec2  d>  sin  d>  ,, 
=  -— (16  + j£--^<fy; 

du  _  sec2  8  sin  8  .  2  sec2  ^  sin  <t>  d<f> 
d$~         5  3  ~dl ' 


*  Let  the  student  not  leave  this  statement  till  he  is  absolutely  con- 
vinced of  its  truth.  An  accurate  figure  on  a  large  scale  will  bring  tbe 
fact  out  clearly. 


TRIGONOMETRIC   FUNCTIONS  123 

On  setting  clu/dd  =  0,  we  obtain  the  equation : 

,.-.  sec2 6  sin 6  _  _  2  sec2 <f> sin  <f>  dcf> 

U  5  3  d0* 

Next,  differentiate  (2) : 

sec-  6d$  +  2  sec2  <£  r/6  =  0, 
or 

(5)  sec20  =  -2sec2<£^. 


Now,  divide  equation  (4)  by  equation  (5)  :  * 

/a\                       sin  6      sin  <f>                      sin  6      5 
(o)  = -£         or         =  -  • 

5  3  sin  <f>     3 

The  result,  stated  in  words,  is  as  follows  :  sin  6  is  to  sin  <f>  as 
the  velocity  in  water  is  to  the  velocity  on  land. 

Let  the  student  work  the  general  problem,  in  which  all  the 
data  are  taken  in  literal  form,  and  verify  the  general  result 
just  stated. 

In  order  actually  to  determine  0,  equations  (2)  and  (6)  must 
be  solved  as  simultaneous  : 

/-.x  f  tan  6  +  2  tan  <£  =  4, 

1  3  sin  6  =  5  sin  <j>. 

This  is  done  best  by  the  method  of  Trial  and  Error,  as  it  is 
called  in  Physics ;  Successive  Approximations  being  the  name 
usually  given  to  it  in  Mathematics.  It  is  a  most  important 
method  in  both  sciences,  and  the  student  should  let  no  oppor- 
tunity go  by  to  use  the  method  whenever,  as  here,  he  meets  a 
case  which  calls  for  it.     Cf.  Chap.  VII,  §  fe. 

The  Corresponding  Problem  in  Optics.  We  have  stated  and 
solved  a  problem  which  is  not  lacking  in  interest,  but  which 
appears  to  have  no  scientific  importance.  This  very  problem, 
however,  occurs  in  Optics.     The  velocity  of  light  is  different  in 

*  i.e.  divide  the  left-hand  side  of  (4)  by  the  left-hand  side  of  (5)  for  a 
new  left-hand  side  ;  and  do  the  same  thing  for  the  right-hand  sides. 


124  CALCULUS 

different  media,  such  as  air  and  water.  Suppose  two  media  to 
be  in  contact  with  each  other,  the  common  boundary  being  a 
plane.  Let  A  be  a  luminous  point,  from  which  rays  emanate 
in  all  directions.  When  the  rays  strike  the  bounding  surface, 
they  are  all  refracted  and  enter  the  second  medium  in  case  the 
velocity  of  light  in  that  medium  is  less  than  in  the  first.  One 
of  the  refracted  rays  will  pass  through  a  given  point  B.  And 
now  the  law  of  light  is  that  the  time  required  for  the  light  to 
pass  from  A  to  B  is  less  for  this  path  than  for  any  other 
possible  path. 

If,  then,  the  velocity  of  light  in  the  first  medium  is  u  *  and 
in  the  second  medium,  v,  we  have : 

sin  6       u 

-— ■  =  -  =  n, 
sin  <f>      v 

where  n  is  the  index  of  refraction  for  the  passage  from  the  first 
medium  into  the  second. 

EXERCISES 

1.  A  wall  27  ft.  high  is  64  ft.  from  a  house.  Find  the  length 
of  the  shortest  ladder  that  will  reach  the  house  if  one  end 
rests  on  the  ground  outside  the  wall. 

Take  the  angle  which  the  ladder  makes  with  the  horizontal 
as  the  independent  variable. 

2.  The  equal  sides  of  an  isosceles  triangles  are  each  8  in. 
long,  the  base  being  variable.  Show  that  the  triangle  of 
maximum  area  is  the  one  which  has  a  right  angle. 

Take  one  of  the  base  angles  as  the  independent  variable,  <£. 

3.  A  gutter  is  to  be  made  out  of  a  long  strip  of  copper 
9  in.  wide  by  bending  the  strip  along  two  lines  parallel  to  the 
edges  and  distant  respectively  3  in.  from  an  edge.  Thus  the 
cross-section  will  be  a  broken  line,  made  up  of  three  straight 
lines,  each  3  in.  long.     How  wide  should  the  gutter  be  at  the 

*  The  letter  u  used  here  has  nothing  to  do  with  the  u  used  above  in 
solving  the  problem. 


TRIGONOMETRIC   FUNCTIONS  125 

top,  in  order  that  its  carrying  capacity  may  be  as  great  as 
possible  ?  Arts.  6  in. 

4.  Johnny  is  to  have  a  piece  of  pie,  the  perimeter  of  which 
is  to  be  12  in.  If  Johnny  may  choose  the  plate  on  which  the 
pie  is  to  be  baked,  what  size  plate  would  he  naturally  select  ? 

5.  A  can-buoy  in  the  form  of  a  double  cone  is  to  be  made 
from  two  equal  circular  iron  plates  by  cutting  out  a  sector 
from  each  plate  and  bending  up  the  plate.  If  the  radius  of 
each  plate  is  a,  find  the  radius  of  the  base  of  the  cone  when 
the  buoy  is  as  large  as  possible.  Ans.   aV-f. 

6.  From  a  circular  piece  of  filter  paper  a  sector  is  to  be  cut 
and  then  bent  into  the  form  of  a  cone  of  revolution.  Show 
that  the  largest  cone  will  be  obtained  if  the  angle  of  the  sector 
is  .8165  of  four  right  angles. 

7.  Two  solid  spheres,  whose  diameters  are  8  in.  and  18  in., 
have  their  centres  35  in.  apart.  At  what  point  in  their  line 
of  centres  and  between  the  spheres  should  a  light  be  placed  in 
order  to  illuminate  the  largest  amount  of  spherical  surface  ? 

Ans.    8  in.  from  the  centre  of  the  smaller  sphere. 

8.  Find  the  most  economical  proportions  for  a  conical  tent. 

9.  A  block  of  stone  is  to  be  drawn  along  the  floor  by  a  rope. 
Find  the  angle  which  the  rope  should  make  with  the  horizontal 
in  order  that  the  tension  may  be  as  small  as  possible. 

Ans.    The  angle  of  friction. 

10.  A  block  of  stone  is  to  be  drawn  up  an  inclined  plane  by 
a  rope.  Find  the  angle  which  the  rope  should  make  with  the 
plane,  in  order  that  the  tension  in  the  rope  be  as  small  as 
possible. 

11.  A  statue  ten  feet  high  stands  on  a  pedestal  that  is  50  ft. 
high.  How  far  ought  a  man  whose  eyes  are  5  ft.  above  the 
ground  to  stand  from  the  pedestal  in  order  that  the  statue  may 
subtend  the  greatest  possible  angle  ? 

12.  A  steel  girder  25  ft.  long  is  moved  on  rollers  along  a 
passageway  12.8  ft.  wide,  and  into  a  corridor  at  right  angles 


126  CALCULUS 

to  the  passageway.  Neglecting  the  horizontal  width  of  the 
girder,  find  how  wide  the  corridor  must  be  in  order  that  the 
girder  may  go  round  the  corner.  .4ns.   5.4  ft. 

13.  A  gutter  whose  cross-section  is  an  arc  of  a  circle  is  to  be 
made  by  bending  into  shape  a  strip  of  copper.  If  the  width 
of  the  strip  is  a,  find  the  radius  of  the  cross-section  when  the 
carrying  capacity  of  the  gutter  is  a  maximum.  Ans.    a/i 

14.  A  long  strip  of  paper  8  in.  wide  is  cut  off  square  at  one 
end.  A  corner  of  this  end  is  folded  over  on  to  the  opposite 
side,  thus  forming  a  triangle.  Find  the  area  of  the  smallest 
triangle  that  can  thus  be  formed. 

15.  In  the  preceding  question,  when  will  the  length  of  the 
crease  be  a  minimum  ? 

16.  The  captain  of  a  man-of-war  saw,  one  dark  night,  a 
privateersman  crossing  his  path  at  right  angles  and  at  a 
distance  ahead  of  c  miles.  The  privateersman  was  making 
a  miles  an  hour,  while  the  man-of-war  could  make  only  b  miles 
in  the  same  time.  The  captain's  only  hope  was  to  cross  the 
track  of  the  privateersman  at  as  short  a  distance  as  possible 
under  his  stern,  and  to  disable  him  by  one  or  two  well-directed 
shots  ;  so  the  ship's  lights  were  put  out  and  her  course  altered 
in  accordance  with  this  plan.  Show  that  the  man-of-war 
crossed  the  privateersman's  track  -  V«2  —  b-  miles  astern  of 
the  latter. 

If  a  =  b,  this  result  is  absurd.     Explain. 

17.  The  illumination  of  a  small  plane  surface  by  a  luminous 
point  is  proportional  to  the  cosine  of  the  angle  between  the 
rays  of  light  and  the  normal  to  the  surface,  and  inversely  pro- 
portional to  the  square  of  the  distance  of  the  luminous  point 
from  the  surface.  At  what  height  on  the  wall  should  an  arc 
light  be  placed  in  order  to  light  most  brightly  a  portion  of 
the  floor  a  ft.  distant  from  the  wall  ? 

Ans.   About  T7^  a  ft.  above  the  floor. 


TRIGONOMETRIC   FUNCTIONS  127 

18.  A  town  A  situated  on  a  straight  river,  and  another  town 
B,  a  miles  farther  down  the  river  and  b  miles  back  from  the 
river,  are  to  be  supplied  with  water  from  the  river  pumped 
by  a  single  station.  The  main  from  the  waterworks  to  A 
will  cost  $  m  per  mile  and  the  main  to  B  will  cost  $  n  per  mile. 
Where  on  the  river-bank  ought  the  pumps  to  be  placed? 

19.  A  telegraph  pole  25  ft.  high  is  to  be  braced  by  a  stay 
20  ft.  long,  one  end  of  the  stay  being  fastened  to  the  pole  and 
the  other  end  to  a  short  stake  driven  into  the  ground.  How 
far  from  the  pole  should  the  stake  be  located,  in  order  that  the 
stay  be  most  effective  ? 

20.  Into  a  full   conical  wine-glass  whose   depth   is   a  and 

generating  angle  a  there  is  carefully  dropped  a  spherical  ball 

of  such  a  size  as  to  cause  the  greatest  overflow.     Show  that  the 

radius  of  the  ball  is 

a  sin  a 


sin  «  +  cos  2  a 


21.  A  foot-ball  field  2  a  ft.  long  and  26  ft.  broad  is  to  be 
surrounded  by  a  running  track  consisting  of  two  straight  sides 
(parallel  to  the  length  of  the  field)  joined  by  semicircular  ends. 
The  track  is  to  be  4  c  ft.  long.  Show  how  it  should  be  made 
in  order  that  the  shortest  distance  between  the  track  and  the 
foot-ball  field  may  be  as  great  as  possible. 

22.*  The  number  of  ems  (or  the  number  of  sq.  cms.  of  text) 
on  this  page  and  the  breadths  of  the  margins  being  given, 
what  ought  the  length  and  breadth  of  the  page  to  be  that  the 
amount  of  paper  used  may  be  as  small  as  possible  ? 

23.  Assuming  that  the  values  of  diamonds  are  proportional, 
other  things  being  equal,  to  the  squares  of  their  weights,  and 
that  a  certain  diamond  which  weighs  one  carat  is  worth  $  m, 
show  that  it  is  safe  to  pay  at  least  $8m  for  two  diamonds 
which  together  weigh  4  carats,  if  they  are  of  the  same  quality 
as  the  one  mentioned. 

*  Exs.  22-25  do  not  involve  Trigonometry. 


128 


CALCULUS 


X  107  ergs 


24.  When  a  voltaic  battery  of  given  electromotive  force 
(E  volts)  and  given  internal  resistance  (r  ohms)  is  used  to 
send  a  steady  current  through  an  external  circuit  of  R  ohms 
resistance,  an  amount  of  work,  W,  equivalent  to 

E2R 
(r  +  Ry 

is  done  each  second  in  the  outside  circuit.  Show  that,  if  dif- 
ferent values  be  given  to  R,  W  will  be  a  maximum  when 
R  =  r. 

25.  An  ice  cream  cone  is  to  hold  one-eighth  of  a  pint.  The 
slant  height  is  I,  and  half  the  angle  at  the  vertex  is  x.  Find 
the  value  of  x  that  will  make  the  cost  of  manufacture  of  the 
cone  a  minimum.  (Ans.   x  =  35°.27.) 


8. 


Let 


Tangents  in  Polar  Coordinates. 

r  =  f(0) 

be  the  equation  of  a  curve  in  polar  coordinates.  We  wish  to 
find  the  direction  of  its  tangent.  The  direction  will  be  known 
if  we  can  determine  the  angle  \p  between  the  radius  vector  pro- 
duced and  the  tangent.     Let  P,  with  the  coordinates  (r0,  00),  be 

an    arbitrary    point 
it  of    the    curve    and 

P':(/-0  +  Ar,  0O  +  A0) 
a  neighboring  point. 
Draw  the  chord 
PP'  and  denote 
the  Z  OP'P  by  if,'. 
Then   obviously 

lim  if/'  =  ij/0. 

To  determine   \p0, 
drop    a    perpendic- 
ular PM  from  P  on 
the  radius  vector  OP'  and  draw  an  arc  PN  of  a  circle  with 
0  as  centre.     The  right  triangle  MP'P  is  a  triangle  of  refer- 


FiG.  45 


TRIGONOMETRIC   FUNCTIONS  129 

ence  for  the  angle  \j/'  and 

Y        MP 
Hence 

P'M 

(1)  cot  \ff0  =  lim  cot  \f/'  =  lim  =-^  ■ 

p'=p  P'=p  MP 

In  the  latter  ratio  we  can  replace  P'M  and  MP  by  more 
convenient  infinitesimals  ;  cf.  Chap.  IV,  §  2.     We  observe  that 

MP  =  r0  sin  A0;         hence         lim  —  =  lim  S11l^  =  1 ; 

Ae^o  roA0      a<h=o     A# 

t'.e.  MP  and  roA0  are  equivalent  infinitesimals. 

Furthermore,  P'M  and  P'N=  Ar  are  also  equivalent  infini- 
tesimals.    For  _,_,      _,,,_       __,_ 
P'M=P'N+  NM 

and 


Hence 


Now,  by  §  3, 


NM  = 

ro~ 

r0  cos  A0. 

NM 

1 

»'o- 

—  cos 
A6> 

A0 

Ar 

Ar 

lim  — - 

-  cos 

Ad 

0. 

A0 


On  the  other  hand,      ,  •     Ar       ,-. 

AfliO  At* 

and  this  quantity  is  not,  in  general,  0.     Hence 

lim™- ft 

A*y>  Ar 

Returning  to  equation  (1)  we  can  now  write  the  last  limit  in 

the  form :  „, , ,  .  ., 

r      P'M     ,.       Ar        1  r. 

lim =  lim =  —  I)gr ; 

p=p  MP      Ae±or0A$      r0 

or,  dropping  subscripts, 

(2)  cot  xf/  =  -Dsr. 


130 


CALCULUS 


In  terms  of  differentials,  this  result  can  be  written  in  either 
of  the  two  forms  : 


(3) 


,  ,       dr               ,       .      rd6 
cot0= ,  tan  i// = 

y     rdd  dr 


Example.     Consider  the  parabola  in  polar  form : 


1  —  cos  <$> 
To  determine  if/.     Here, 

,   _       m  sin  <fr  d(f> 

(l-cos<£)2' 
Hence 

f  /  _       msin<£d<£       1  —  cos 

(1  —  cos  <f>)~  md(f> 

sin<£ 


Fig.  46 


1  —  COS  <f> 

In  particular,  at  the  extremity 
of  the  latus  rectum,  we  have  : 


cot  01      _  =  — 1, 


^  =  I  +  7r' 


and  thus  we  obtain  anew  the  result   that  the  tangent  there 
makes  an  angle  of  45°  with  the  axis  of  the  parabola. 
Again,  at  the  vertex, 


cot  I 


=  0, 


* 


and  the  tangent  there  is  verified  as  perpendicular  to  the  axis. 

From  the  above  equation, 

,  ,  sin  <A 

cot  if/  = ~ — , 

1  —  cos  <f> 

a  simple  relation  between  0  and  <f>  can  be  deduced.     Since 


sin<f> 
1  —  cos  </> 


2  sin*  cos* 

2_cot^ 

2  sin'*  2' 


TRIGONOMETRIC   FUNCTIONS  131 

it  follows  that  ,  ,  ,  d> 

cot  \b  =  —  cot  -£-  • 
Y  2 

But,  for  any  angle,  x, 

cot  (tt  —  x)  —  —  cot  x. 
Setting  x  =  ip  in  the  above  equation,  we  have : 

cot  (jr  —  ij/)—  cot*- 

Hence  *  .      <b 

r      2' 

or,  ?^e  supplement  of  \p  is  equal  to  p.     Thus  we  have  a  new 

proof  of  the  familiar  property  of  the  parabola,  that  the  tangent 
at  any  point  P  of  the  curve  bisects  the  angle  between  the  focal 
radius,  OP,  and  a  parallel  to  the  axis,  drawn  through  P. 

EXERCISES 

1.  Plot  the  spiral,  r  =  6, 

and  show  that  the  angle  at  which  it  crosses  the  prime  direction 
whenO  =  2Tr  is   80°  57'. 

2.  Plot  the  spiral,  _  _  1 

Show  that  it  has  an  asymptote  parallel  to  the  prime  vector. 

Suggestion.  Consider  the  distance  of  a  point  P  of  the  curve 
:rom  the  prime  direction,  and  find  the  limit  of  this  distance 
.vhen  9  approaches  0. 

Determine  the  angle  at  which  the  radius  vector  correspond- 
ug  to  6  =  7r/2  meets  this  curve. 

3.  Plot  the  cardioid, 

r  =  a  (1  —  cos  <f>), 

*  The    trigonometric    equation    admits    a  second   solution,    namely, 
■""  —  i/O  +  t  =.0/2.     if^  however,  we  agree  to  take   0   and   \j/   so   that 
_■  0  <  2  it   and   0  _i  0  <  71- ,  this  second  solution  is  ruled  out. 


132  CALCULUS 

and  show  that  .  .  sin  <4 

COt  {{/  = 


1  —  cos  <f> 

At  what  angle  is  the  curve  cut  by  a  line  through  the  cusp 
perpendicular  to  the  axis  ? 

4.  Prove  that,  for  the  cardioid, 

y      2 

5.  Show  that  the  tangent  to  the  cardioid  is  parallel  to  the 
axis  of  the  curve  when  <f>  =  ^tt. 

6.  At  what  points  of  the  cardioid  is  the  tangent  perpen- 
dicular to  the  axis  of  the  curve  ? 

7.  Determine  the  rectangle  which  circumscribes  the  cardioid 
and  has  two  of  its  sides  parallel  to  the  axis  of  the  curve. 

8.  Show  that,  for  the  lemniscate, 

?-2  =  a2  cos  26, 
the  angle  ^  is  given  by  the  equation : 
cot^  =  —  tan  20. 

Hence,  show  that  ,      rr  ,  o  a 

^  =  2 

9.  At  what  points  of  the  lemniscate  is  the  tangent  parallel 
to  the  axis  *  of  the  curve  ? 

Ans.  At  the  point  for  which  6  =  tt/6,  and  the  points  which 
correspond  to  it  by  symmetry. 

10.  The  points  of  the  curve 

r  =/(*), 

at  which  the  tangent  is  parallel  to  the  prime  vector,  are  ev*j 

dently  those  for  which 

J  y  =  r  sin  <f>, 

*  The  axis  of  any  curve  is  a  line  of  symmetry.  The  lemniscate  haj 
two  such  lines.  The  axis  referred  to  in  the  text  is  that  one  of  these  lines 
which  passes  through  the  vertices  of  the  curve. 


TRIGONOMETRIC   FUNCTIONS  133 

considered  as  a  function  of  <f>  through  the  mediation  of  the 
equation  of  the  curve,  has  a  maximum,  a  minimum,  or  a  certain 
point  of  inflection.     For  these  points,  then, 

dy  ,    ,     ■     ,  dr      A 

-^-  =  r  cos  <£  +  sm  <f>  —  =  0. 
dcf>  dcf> 

Show  that  this  condition  is  equivalent  to  the  one  used  above 
in  the  special  cases  considered,  namely : 

\jj  +    <f>  =   IT. 

11.  Plot  the  curve,       r  =  a  cos  26, 
taking  a  =  5  cm.     Show  that  for  this  curve 

cot  iff  =  —  2  tan  26. 

12.  At  what  points  of  the  curve  of  question  11  is  the  tan- 
gent parallel  to  the  axis  ? 

Ans.    For  one  of  the  points,  tan  6  =  — -  • 

V5 

13.  Plot  the  curve,       r  =  a  cos  30, 
taking  a  =  5  cm.     Show  that 

cot^  =  —  3  tan  30. 

14.  At  what  points  of  the  curve  of  question  13  is  the  tan- 
gent parallel  to  the  axis  of  the  lobe  ? 

Ans.    For  one  of  these  points,  tan  6  =\  1  H 

V        V3 

15.  The  equation        ^  _         m 

1  +  sin  <£ 

represents  a  parabola  referred  to  its  focus  as  pole.  Give  a 
direct  proof  that  the  tangent  to  this  curve  at  any  point  bisects 
the  angle  formed  by  the  focal  radius  drawn  to  this  point  and 
a  parallel  to  the  axis  through  the  point. 

16.  Show  that  the  tangent  to  the  hyperbola 

m 
r  = 

1  —  V3  cos  cf> 


134  CALCULUS 

at  the  extremity  of  the  latus  rectum  makes  an  angle  of  60° 
with  the  transverse  axis. 

17.   Prove  that  the  tangent  to  the  ellipse 

r  =  — JS 

V3  —  cos  (j> 

at  the  extremity  of  the  latus  rectum  makes  an  angle  of  30° 
with  the  major  axis. 

9.    Differential  of  Arc.     Let 

(1)  </=/(*) 

be  the  equation  of  a  given  curve.  Let  P,  with  the  coordinates 
(x,  y),  be  a  variable  point,  and  A  a  fixed  point  of  the  curve. 
Denote  the  length  of  the  arc  AP  by  s.  Then  s  is  a  function 
of  x ;  for,  when  x  is  given,  we  know  P  and  thus  s. 

It  is  possible  to  determine  the  derivative  of  s,  Dxs,  as  fol- 
lows.   By  the  Pythagorean  Theorem  we  have  (Chap.  IV,  Fig.  33), 

PP'2  =  Ax2  +  Ay\ 

Let  Ax  approach  0  as  its  limit.     Then 

lim  (*fy  =  1  +  lim  fMV  =  1  +  (p^y. 

Since  by  §  3  the  chord  PP'  and  the  arc  PP'=  As  are 
equivalent  infinitesimals,  it  follows  from  the  Fundamental 
Theorem  of  Chap.  IV,  §  2  that,  in  the  above  equation,  PP' 
can  be  replaced  by  As.     Hence 

±z±o\  Ax  J       ±x±o\AxJ 
and  consequently 

(2)  (Dxsf=l +(i^,y. 


TRIGONOMETRIC   FUNCTIONS  135 


On  replacing  the  derivatives  in  (2)  by  their  values  in  terms 
of  differentials,  we  have 

or  \dxj  \dxj 

(3)  ds-  =  dx2  +  dy\ 

This  formula  is  easily  interpreted  geometrically  by  means 
of  the  triangle  PMQ,  Fig.  33.     Since 

PM=  dx        and         MQ  =  dy, 

it  follows  from  the  Pythagorean  Theorem  that 

(4)  PQ  =  ds. 

It  is  obvious  geometrically  that  ds  and  As  differ  from  each 
other  by  an  infinitesimal  of  higher  order;  i.e.  that  they  are 
equivalent  infinitesimals.* 

Formulas  for  sin  t,  cos  t.  From  the  triangle  PMQ  we  can 
write  down  two  further  formulas  : 

/crx  dy  dx 

(5)  sin  t  =  — ,  cos  t  =  — 
v  ;                                      ds'  ds 

These  formulas  presuppose  a  suitable  choice  of  t.  As  s  in- 
creases, the  point  P  describes  the  curve  in  a  definite  sense. 
Let  this  be  chosen  as  the  positive  sense  of  the  tangent  line  at 
P.  Then  t  shall  be  the  angle  between  the  positive  axis  of  x 
and  this  line.  If  t  were  taken  as  the  angle  which  the  op- 
positely directed  tangent  makes  with  the  positive  axis  of  x, 
the  —  sign  must  be  written  before  each  right-hand  side  in  (5). 

The  formulas  (5)  suggest  that  x  and  y  can  be  taken  as  func- 
tions of  s : 

x=<f>(s),  y  =  ns). 

*  In  case  the  coordinates  x  and  y  are  expressed  as  functions  of  a  third 
variable  t,  dx  will  not  in  general  be  equal  to  Ax,  but  will  differ  from  it  by 
an  infinitesimal  of  higher  order.  The  triangle  PMQ  will  then  be  replaced 
by  a  similar  triangle  PM x  Qu  in  which  Mi  lies  on  the  line  PM,  its  distance 
from  M  being  an  infinitesimal  of  higher  order. 


136  CALCULUS 

This  is,  of  course,  always  possible,  since,  when  s  is  given,  P, 
and  hence  also  x  and  y,  are  determined. 
Since 


(6)  <is  =  ±  vdx1  +  dy  , 

we  have  from  (5) 

,ns         ■  'hi  dx 

(t)        smi-  =  ± —      "         ,  cost  =  ± 


^/dx-  +  dy1  V(fa2  +  dy'1 

no  matter  what  choices  of  s  and  r  are  made.*     Furthermore, 

dy 

/os  dx 

(8)        8iHT  =  i —  -,  cost  =  ± 


Which  sign  is  to  be  used  in  (8)  depends  on  which  of  the  two 
possible  determinations  has  been  chosen  for  t.  Thus  t  in  a 
given  case  might  be  30°  or  30°  +  180°  =  210°.  If  the  first 
choice  were  made,  r  =  30°,  then  sin  t,  cos  t,  and  dy/dx  =  tan  r 
would  all  be  positive  quantities,  and  hence  the  upper  signs 
must  be  taken.  But  if  the  other  choice,  t  =  210°,  is  made,  then 
sin  t  and  cos  t  are  negative,  and  the  lower  signs  hold. 

Example.     Consider  the  parabola 


Let  Pliea  point  of  the  curve  which  lies  in  the  first  quadrant. 

Since 

tan  t  =  -^  =  2x 
dx 

is  here  positive,  t  may  be  taken  as  an  angle  of  the  first  quad- 
rant.    In  that  case,  formulas  (8)  give 

2x  1 

Sill  T  =  ,  COS  T  = 


Vl  +  4cc2  Vl  +  4x2 

I  in'  signs  in  (6)  and  (7)  are  not  necessarily  the  same  ;  also  in  (7)  and  (8). 


TRIGONOMETRIC   FUNCTIONS  137 

If  P  is  a  point  of  the  curve  which  lies  in  th,e  second  quad- 
rant, tan  t  is  negative,  and  t  is  an  angle  of  the  second  or  fourth 
quadrant.     If  we  choose  to  take  t  as  an  angle  of  the  second  . 
quadrant,  formulas  (8)  become 

2x  1 

S1UT  = — ,  COSt  = 


Vl+4cc2  Vl+4a;2 

We  may,  however,  equally  well  take  t  as  an  angle  of  the  fourth 
quadrant.     Then 

2x  1 

COS  T  = 


Vl  +  4x-2  Vl  +  4z2 

In  each  case,  one  of  the  numbers,  sin  t  and  cos  t,  is  positive, 
the  other,  negative. 

Polar  Coordinates.  Similar  considerations  in  the  case  of 
the  curve  .... 

lead  to  the  following  formulas  ;  cf .  Fig.  45  : 
PP'2  =  P'M2  +  MP2. 

Hence  lim  f^Y=  lim  (££)'+  lim  (^ 

±e±o  \  A<9  J       se=o  \  A0  J       a<m=o  \  A<9 

Now,  the  chord  PP'  and  the  arc  PP'  =  As  are  equivalent 
infinitesimals.  Moreover,  P'M  and  Ar  are  equivalent;  and 
MP  and  r0A6  are  equivalent.     Hence 

(ZV)2=OV)2-t-n>2. 

Dropping  the  subscript  and  writing  the  derivatives  in  terms 
of  differentials  we  have,  then  : 


<9>  tSMs)+"' 

or 

(10)  ds*  =  dr2  +  r2d9\ 


138  CALCULUS 

Furthermore, 

/-mn                         •     ,      rdO  ,      dr 

(11)  sin  \p  = ,  cos  ii  =  — , 

ds  ds 

the  tangent  PT  being  drawn  in  the  direction  of  the  increasing 
s,  and  i/>  being  taken  as  the  angle  from  the  radius  vector  pro- 
duced to  the  positive  tangent. 

10.  Rates  and  Velocities.  The  principles  of  velocities  and 
rates  were  treated  in  Chapter  III,  §  8.  We  are  now  in  a 
position  to  deal  with  a  wider  range  of  problems. 

We  note  the  following  formulas.  Let  a  point  P  describe 
the  curve  y=f(x) 

Let  s  denote  the  length  of  the  arc,  measured  from  an  arbitrary 
point  in  an  arbitrary  sense,  and  let  t  be  the  angle  from  the 
positive  direction  of  the  axis  of  x  -to  the  tangent  at  P  drawn 
in  the  direction  of  the  increasing  arc.  Then  the  components 
of  the  velocity  (v  =  ds/dt)  of  P  along  the  axes  are,  respectively  : 

....  dx  dy 

( 1 )  —  =  v  cos  t,  -S.  =  v  sm  t. 
v  ;                         dt  dt 

Let  a  point  P  describe  the  curve 

(2)  r=F(9). 

Let  s  denote  the  length  of  the  are,  measured  from  an  arbitrary 
point  in  an  arbitrary  sense ;  and  let  if/  be  the  angle  from  the 
radius  vector,  produced  beyond  P,  to  the  tangent  at  P  drawn 
in  the  direction  of  the  increasing  arc.  Then  the  components 
of  the  velocity  (v  =  ds/dt)  of  P  along  the  radius  vector  pro- 
duced and  perpendicular  to  the  same  (the  sense  of  the  increas- 
ing $  being  taken  as  positive  for  the  latter)  are  respectively  : 

/o\  dr  .  d6  , 

(3)  —  =  v  cos  ip,         r  —  =  u  sm  \b. 

w  dt  Y  dt  Y 

Example  1.  A  railroad  train  is  running  at  the  rate  of  30 
miles  an  hour  along  a  curve  in  the  form  of  a  parabola  : 

f-=  1000  a-, 


TRIGONOMETRIC   FUNCTIONS  139 

the  axis  of  the  parabola  being  east  and  west,  and  the  foot  being 
taken  as  the  unit  of  length.  The  sun  is  just'  rising  in  the  east. 
Find  how  fast  the  shadow  of  the  locomotive  is  moving  along 
the  wall  of  the  station,  which  is  north  and  south,  when  the 
distance  of  the  shadow  from  the  axis  of  the  parabola  is  300  ft. 
The  first  thing  to  do  is  to  draw  a  suitable  figure,  introduce 
suitable  variables,  and  set  down  all  the  data  not  already  put 
into  evidence  by  the  figure.  Thus  in  the 
present  case  we  have,  in  addition  to  the  ac- 
companying figure,  the  further  data :  (a)  the 
velocity  of  the  train  ;  this  must  be  expressed 
in  feet  per  second,  since  we  wish  to  retain 

the  foot  as  the  unit  of  length  for  the  equa- 

.  Fig.  47 

tion  of  the  curve.     Now,  30  miles  an  hour 

is  equivalent-  to  44  feet  a  second.     On  the  other  hand,  another 

expression    for   the   velocity   is  ds/dt.     Hence   we   have,  on 

equating  these  two  values, 

^  =  44. 

dt 

(b)  We  must  set  down  explicitly  at  this  point  the  equation 

of  the  curve, 

yt=  1000  x. 

To  sum  up,  then,  we  first  draw  the  figure  and  then  write 
down  the  supplementary  data  : 

ds 
Given  a)  —  =  44, 

7  dt 

and  b)  y*  =  1000  a. 

The  second  thing  to  do  is  to  make  clear  what  the  problem 
is.     In  the  present  case  it  can  be  epitomized  as  follows : 

To  find  (*»)       • 

We  are  now  ready  to  consider  what  methods  are  at  our  dis- 
posal for  solving  the  problem.     We  observe  that  ds  occurs  in 


140  CALCULUS 

the  data.  Obviously,  then,  we  must  make  use  of  the  one  gen- 
eral theorem  we  Mow  which  gives  an  expression  for  ds  when 
the  equation  of  the  curve  comes  to  us  in  Cartesian  coordinates, 
—  namely,  the  theorem  : 

ds2  =  dx2  +  dy2. 

Since  dx  occurs  neither  in  the  data  nor  in  the  conclusion, 
we  wish  to  eliminate  it.  This  can  be  done  by  means  of  the 
equation  of  the  path  b).     Differentiating  b)  we  have: 

2v/(?//  =  1000rt.i\ 

Hence  dx  =  •'  "  • 

500 

y2  di/2 
Consequent  1  y  ds2  =  ,y     •    4-  dy2 


and  ds=yj-gL^-  +  ldy. 

The  next  step  is  obvious ;  divide  through  by  dt : 

—  =  \  I    !/1     4- 1  ^ 
dt     \  5002         dt  ' 

In  this  last  equation,  replace  ds/dt  by  its  known  value  from 
a),  and  we  now  have  an  equation  for  determining  dy/dt : 

dy  _         44 

\  5002 

Finally,  bring  into  action  the  particular  value  of  y  with 
which  alone  the  proposed  equation  is  concerned,  namely, 
y  =  300 : 

\dtjy=m     V.624-l      Vl.36 

or,  the  rate  at  which  the  shadow  is  moving  along  the  wall  of 
the  station  is  37.73  ft.  a  second. 


TRIGONOMETRIC   FUNCTIONS  141 

Angular  Velocity.  By  the  angular  velocity,  w,  with  which  a 
line  is  turning  in  a  given  plane  is  meant  the  rate  at  which  the 
angle,  <f>,  made  by  the  rotating  line  with  a  fixed  line,  is  in- 
creasing : 

d<f> 
(a  =  — • 
dt 

Example  2.     A  point  is  describing  the  cardioid 
r  =  a  (1  —  cos  6) 

at   the  rate  of   c  ft.  a  second.     Find  the  rate  at  which'  the 
radius  vector  drawn  to  the  point  is  turning  when  6  =  tt/2. 
The  formulation  of  this  problem  is  as  follows  : 


Given 

«) 

ll.N 

dt~ 

and* 

b) 

r=  a 

To  find 

fd$ 

[dt, 

Since, 

from  §  9  (10), 

,  -         ..  ds2  =  d,r*  +  r2d02, 

and  from  o), 

,  £  ,,         ,,  dr  =  a  sin  6  dd, 

it  follows  that 

ds2  =  a2  sin2  6  dO2  +  a-(l  -  cos  Of  d& 

=  a2  dd2  [sin2  6  +  1  -  2  cos  6  +  cos-  ff] 

=2  aW  •  (1  -  cos  6)  =  4  a2  sfti-  £  r702. 

Li 

Hence,  s  being  measured  from  the  cusp, 

ds  =  2asin-d0, 
2 

,  ds     0      ■    0d6 

and  —  =  2  a  sin  -  — 

d*  2d« 

*  The  student  should  make  a  free-hand  drawing  of  the  curve. 


142  CALCULUS 

Consequently,  by  the  aid  of  a) 

dO  c 


dt      0      ■    6' 
2  a  sm  - 

2 

and  thus,  finally  (f^\  c 

Vlt)e=?_~ay/2' 

EXERCISES 

1.  A  point  describes  a  circle  of  radius  200  ft.  at  the  rate 
of  20  ft.  a  second.  How  fast  is  its  projection  on  a  fixed 
diameter  travelling  when  the  distance  of  the  point  from  the 
diameter  is  100  ft.  ?  Ans.    10  ft.  a  second. 

2.  A  flywheel  15  ft.  in  diameter  is  making  3  revolutions  a 
second.  The  sun  casts  horizontal  rays  which  lie  in  or. are 
parallel  to  the  plane  of  the  flywheel.  A  small  protuberance 
on  the  rim  of  the  wheel  throws  a  shadow  on  a  vertical  wall. 
How  fast  is  the  shadow  moving  when  it  is  4  ft.  above  the 
level  of  the  axle  ? 

3.  A  revolving  light  sends  out  a  bundle  of  rays  that  are 
approximately  parallel,  its  distance  from  the  shore,  which  is 
a  straight  beach,  being  half  a  mile,  and  it  makes  one  revolu- 
tion in  a  minute.  Find  how  fast  the  light  is  travelling  along 
the  beach  when  at  the  distance  of  a  quarter  of  a  mile  from  the 
nearest  point  of  the  beach. 

4.  A  point  moves  along  the  curve  r  =  l/6  at  the  rate  of 
<>  ft.  a  second.  How  fast  is  the  radius  vector  turning  when 
0  =  2tt? 

5.  In  the  example  of  the  ladder,  Chap.  Ill,  §  8,  Ex.  5,  find 
how  fast  the  ladder  is  rotating  at  the  instant  in  question. 

6.  At  what  rate  is  the  direction  of  the  second  ship  from  the 
first  changing  at  the  instant  in  question,  in  Ex.  2  of  Chap.  Ill, 
§8? 


TRIGONOMETRIC   FUNCTIONS 


143 


7.  How  fast  is  the  direction  of  the  man  from  the  lamp- 
post changing  in  Ex.  12  of  Chap.  Ill,  §  8  ? 

8.  The  sun  is  just  setting  as  a  baseball  is  thrown  vertically 
upward  so  that  its  shadow  mounts  to  the  highest  point  of  the 
dome  of  an  observatory.  The  dome  is  50  ft.  in  diameter. 
Find  how  fast  the  shadow  of  the  ball  is  moving  along  the 
dome  one  second  after  it  begins  to  fall,  and  also  how  fast  it  is 
moving  just  after  it  begins  to  fall. 

9.  Let  AB,  Fig.  48,  represent  the   rod  that   connects   the 


piston  of  a  stationary  engine  with  the  fly-wheel 
the  velocity  of  A  in  its  rectilinear  path, 
and  v  that  of  B  in  its  circular  path, 
show  that 


If  u  denotes 


Fig.  48 


Fig.  49 


u  =(sin  6  +  cos  8  tan  <fi)v. 

10.    Find  the  velocity  of  the  piston  of 
a  locomotive  when  the  speed  of  the  axle  of  the  drivers  is  given. 

11.  A  drawbridge  30  ft. 
long  is  being  slowly  raised 
by  chains  passing  overa  wind- 
lass and  being  drawn  in  at 
the  rate  of  8  ft.  a  minute.  A 
distant  electric  light  sends 
out  horizontal  rays  and  the 
bridge  thus  casts  a  shadow 
on  a  vertical  wall,  consisting  of  the  other  half  of  the  bridge, 
which  has  been  already  raised.  Find  how  fast  the  shadow 
is  creeping  up  the  wall  when  half  the  chain  has  been 
drawn  in. 

12.  A  man  walks  across  the  floor  of  a  semicircular  rotunda 
100  ft.  in  diameter,  his  speed  being  4  ft.  a  second,  and  his 
path  the  radius  perpendicular  to  the  diameter  joining  the 
extremities  of  the  semicircle.  There  is  a  light  at  one  of  the 
latter  points.  Find  how  fast  the  man's  shadow  is  moving  along 
the  wall  of  the  rotunda  when  he  is  halfway  across. 


144  CALCULUS 

13.  A  man  in  a  train  that  is  running  at  full  speed  looks  out 
of  the  window  in  a  direction  perpendicular  to  the  track.  If 
he  fixes  his  attention  successively  for  short  intervals  of  time 
on  objects  at  different  distances  from  the  train,  show  that  the 
rate  at  which  he  has  to  turn  his  eyes  to  follow  a  given  object 
is  inversely  proportional  to  its  distance  from  him. 

14.  Water  is  flowing  out  of  a  vessel  of  the  form  of  an  in- 
verted cone,  whose  semi-vertical  angle  is  30°,  at  the  rate  of  a 
quart  in  2  minutes,  the  opening  being  at  the  vertex.  How 
fast  is  the  level  of  the  water  falling  when  there  are  4  qt.  of 
water  still  in  ? 

15.  Suppose  that  the  locomotive  of  the  first  of  the  Examples 
worked  in  the  text  is  approaching  the  station  at  night  at  the 
rate  of  20  miles  an  hour,  its  headlight  sending  out  a  bundle 
of  parallel  rays.  How  fast  will  the  spot  of  light  be  moving 
along  the  wall  of  the  station  when  the  distance  of  the  head- 
light from  the  vertex  A  of  the  parabola,  measured  in  a  straight 
line,  is  500  ft.  ? 

Assume  that  the  wall  is  perpendicular  to  the  axis  of  the 
parabola  and  distant  75  ft.  from  the  vertex. 

16.  In  the  preceding  question,  how  fast  will  the  bundle  of 
lays  be  rotating? 

17.  A  point  describes  a  circle  with  constant  velocity.  Show 
that  the  velocity  with  which  its  projection  moves  along  a  given 
diameter  is  proportional  to  the  distance  of  the  point  from  this 
diameter. 

18.  A  point  P  describes  the  arc  of  the  ellipse 

9z2  +  4?/2=36, 

which  lies  in  the  first  quadrant,  at  the  rate  of  12  ft.  a  second. 
The  tangent  at  P  cuts  off  a  right  triangle  from  the  first  quad- 
rant. How  fast  is  the  area  of  this  triangle  changing  when  P 
passes  through  the  extremity  of  the  latus  rectum  ?  Is  the  area 
increasing  or  decreasing  ? 


TRIGONOMETRIC    FUNCTIONS  145 

19.  A  point  P  describes  the  cardioid 

r  =  5  (1  —  cos  6) 

at  the  rate  of  12  crn.  a  second.     The  tangent  at  P  cuts  the 
axis  of  the  curve  in  Q.     How  fast  is  Q  moving  when  6  =  tt/2  ? 

20.  The  sun  is  just  setting  in  the  west  as  a  horse  is  running 
around  an  elliptical  track  at  the  rate  of  m  miles  an  hour.  The 
axis  of  the  ellipse  lies  in  the  meridian.  Find  the  rate  at 
which  the  horse's  shadow  moves  on  a  fence  beyond  the  track 
and  parallel  to  the  axis. 


CHAPTER   VI 

LOGARITHMS   AND   EXPONENTIALS 

1.  Logarithms.  The  logarithms  with  which  the  student  is 
familiar  are  those  which  are  ordinarily  used  for  computation. 
The  base  is  10,  and  the  definition  of  log10  x  is  as  follows : 

?/  =  log10  x        if         10^  =  x. 

These  are  called  denary,  or  Briggs's,  or  common  logarithms. 

More  generally,  any  positive  number,  a,  except  unity,  can 
be  taken  as  the  base,  the  definition  of  logtt  x  then  being : 

(1)  y  =  logax        if        ay=x. 

y  =  log  x 


Fig.  50 


The  accompanying  figure  represents  in  character  the  graph 
of  the  function  log0  x  for  any  a  >  1.     It  is  drawn  to  scale  for 

146 


LOGARITHMS  AND   EXPONENTIALS  147 

a  =  2.71828.  The  reason  for  this  choice  of  a  will  appear 
shortly. 

From  the  definition  it  follows  at  once  that 

(2)  logol  =  0,  log„a  =  l. 

Only  positive  numbers  have  logarithms.  For,  av  is  always 
positive.  Hence,  if  x  be  given  a  negative  value  (or  the  value 
0),  the  second  equation  under  (1)  above  cannot  be  satisfied  by 
any  value  of  y. 

The  two  leading  properties  of  logarithms  are  expressed  by 
the  equations :  * 

(I)  logP+logQ  =  log(PQ) 

(II)  log  P*  =  n  log  P. 

Here,  P  and  Q  are  any  two  positive  numbers  whatever,  and  n 
is  any  number,  positive,  negative,  or  zero.  The  base,  a,  is 
arbitrary.     Thus 

log  10  =  log  2  +  log  5 

and  log  V7  =  log  75  =  \  log  7. 

From  equation  (I)  it  follows  that 

(3)  logl  =  -logQ 

and 

p 

(4)  log  -  =  log  P  -  log  Q. 


For,  if  we  set  P  =  1/Q  in  (I),  we  have 
But,  by  (2), 


log  1  =  log  -  +  log  Q. 


log  1  =  0. 

*  The  student  should  recall  the  proofs  of  these  theorems,  which  he 
1  learned  in  the  earlier  study  of  logarithms,  and  make  sure  that  he  can 
reproduce  them.  Proofs  of  the  theorems  are  given  in  the  author's 
Differential  and  Integral  Calculus,  p.  76. 


148  CALCULUS 

Hence  ,      1  ,      r.  , 

S0  =  ~    g^'  q       ' 

Again,  write  (1)  in  the  form 

log(PQO  =  logP+logV, 

and  now  set  Q'  =  1/Q.     Then 

h>g-  =  log7J+  log  -• 
But  log  —  =  —  log  Q. 


Hence 


Q 
p 

log  -  =  log  P  -  log  Q,  q.  e.  d. 

For  example,  * 

log  (a  +  b)  -  log  a  =  log  (\  +  _\ 

as  we  see  by  setting,  in  equation  (4), 

P  =  a  +  b,  Q  =  a. 

As  a  further  example  of  the  application  of  equation  (II)  we 
may  cite  the  following  : 

log(a  +  6)  =  logKa  +  6)^ 

For,  if  P  =  a  -+■  b   and    n  =  - ,  the  left-hand  side  of  this  equa- 
tion has  the  value  n  log  P. 

A  Further  Property  of  Logarithms.  When  it  is  desired  to 
express  a  logarithm  given  to  a  certain  base,  a,  in  terms  of 
logarithms  taken  to  a  second  base,  b,  the  following  relation  is 
needed : 

(III)  logaa-  =  -^^. 

log„a 

The  proof  of  (III)  is  as  follows.     Let 

y  =  loga  x,  a"  =  x. 


LOGARITHMS   AND    EXPONENTIALS  149 

Take  the  logarithm  of  each  side  of  this  equation  to  the  base  b : 
(5)  log,  a"  =  log6  x. 

But  the  left-hand  side  can  be  transformed  by  (II),  if  in  (II) 
we  take  b  as  the  base,  thus  having 

log6  Pn  =  n  log,  P. 
Here,  let 

Then  log6  a"  =  y  log6  a, 

and  (5)  now  becomes  : 

y  log6  a  =  log,  x. 


Hence 


y  =  |5&£?,         or         log^  =  ^,  q.e.d. 

log,  a  log,  a 


Example.     Let  b  =  10  and   let   a  =  2.718.     To   find   loga  2. 
From  (III), 

logo2=     logxo2     =.3010  =  .6932. 
Sa         log10  2.718      .4343 

7\«o  Identities.     Just  as,  for  example, 

^x3  =  x        and         (va;)3  =  £, 

i    no  matter  what  value  x  may  have,  so  we  can  state  two  identi- 
i    ties  for  logarithms  and  exponentials:     In  the  second  equation 

(1),  replace  y  by  its  value  from  the  first  equation.     Thus  the 

equation 

(6)  aXos«x  =  x 

is  seen  to  hold  for  all  positive  values  of  x. 

Secondly,  replace  x  in  the  first  equation  (1)  by  its  value 
from  the  second  equation  : 

y  =  loga  ay. 

We  can  equally  well  write  <k  instead   of  y,  understanding 
now    by  x  any   number   whatever,  and   we   have,   then,    the 


150  CALCULUS 

identity 

(7)  loga  ax  =  x. 

This  equation  holds  for  all  values  of  x,  positive,  negative,  or 
zero. 

EXERCISES 

1.  Show  that 

logio  -°9o0  =  —  .0482. 

2.  Kind  log10  .09420.  Arts.    -  1.0259. 

3.  Compute  2.718-5642.  Ans.    1.758. 

4.  Compute  2.718--8710.  Ans.   0.4186.  . 

5.  Compute  irn.  6.    Compute  V2  ®. 

7.  Show  that 

log  tan  9  =  log  sin  9  —  log  cos  0,  0  <  9  <  -  • 

8.  Show  that 

log  sin  6  +  log  cos  0  =  log  S1"  2  ^ ,       0  <  9  <  -  ■ 

9.  Show  that 

,      1  —  eos  9      o ,        •    9  a  ^  />   ^  o 

log        -     -  =  21ogsin-,  0  <  6  <  2tt. 

10.  If  (x,  y)  are  the  Cartesian  coordinates  of  a  point  distinct 

from  the  origin,  and   (r,  9)  the  polar  coordinates  of  the  same. 

point,  show  that         .  ,  .      ,  „ 

log  r  =  \  log  (x2  +  i/1). 

11.  Prove  that 

log  (a2  —  b2)  =  log  (a  +  b)  +  log  (a  —  b), 
provided  a  +  b  and  a  —  b  are  both  positive  quantities. 

12.  Simplify  the  expression 

log  (1  + a:6)- log  (1  +  x2). 

13.  Show  that 

V(ez  —  e~xy  +  4  =  ez  +  e~x, 

where  e  has  the  value  2.7182. 


LOGARITHMS   AND   EXPONENTIALS  151 

14.    Simplify  the  expression 


15.    Show  that 

ilog  (1  +  0'=  log  (1 +*)'". 

2.    Differentiation  of  Logarithms.     In  order  to  differentiate 

the  function  . 

V  =  logu  x, 

it  is  necessary  to  go  back  to  the  definition  of  a  derivative, 
Chap.  II,  §  1,  and  carry  through  the  process  step  by  step. 

Give  to  x  an  arbitrary  positive  value,  x0,  and  compute  the 
corresponding  value,  y0,  of  the  function : 

(1)  Vo  =  log3  x0. 

Next,  give  to  x  an  increment  Ax  (subject  merely  to  the  restric- 
tion that  x0  +  Ace  is  positive  and  Ax  =£  0)  and  compute  the  new 
value,  y0  +  A?/,  of  the  function  : 

(2)  y0  +  A?/  =  loga  O0  -f  Ax). 

From  (1)  and  (2)  it  follows  that 

Ay  _  loga  (x0  +  Ax)  —  log0  x0 

Ax  Ax 

It  is  at  this  point  that  the  specific  properties  of  the  loga- 
rithmic function  come  into  play  for  the  purpose  of  transform- 
ing the  last  expression.     By  §  1,  (4), 

loga  0*b  +  Az)  -  loga  x0  =  loga  M  +  —  J, 
and  hence 

(3)  ^  =  ±logA+*? 

Ax      Ax         \         x0 

We  next  replace  the  variable  Ax  by  a  new  variable  t  as 
follows  : 


152  CALCULUS 


t  =  —  or  Ax  =  xnt. 


x0 
Thus  (3)  takes  on  the  form 


^=ilog„(l+0=i 


\  log.  (1  +  0 


From  (II),  §  1,  the  bracket  is  seen  to  have  the  value 

i 

loga(l  +  0r. 
and  hence 

(4)  %>  =  ±loga(l+t)\ 

A.i'       Xq 

As  Ax  approaches  0  as  its  limit,  t  also  approaches  0,  and  so 

(5)  lim  ^  =  -  lim  loga  (1  +  t)  K 
Ax^oAa;      Xq  e=o 

Now,  the  variable  (1  -f  t)'  approaches  a  limit  when  t  ap- 
proaches 0,  and  this  limit  is  the  number  which  is  represented 
in  mathematics  by  the  letter  e;  cf.  §  3.  Its  value  to  five 
places  of  decimals  is 

e  =  2.71828...  ; 

cf.  §  3.     Moreover,  log  x  is  a  continuous  function  of  x,  as  is 
shown  in  a  detailed  study  of  this  function.*     Hence 


lim  logu  (1  +  0 '  =  loga     lim  (1  +  0 (     =  logo  e. 

On  substituting  this  value  in  the  right-hand  side  of  (5)  we 
have : 

n         log.  e 

Dxy  =  -§s_ , 

x0 

*  Such  a  treatment  is  too  advanced  to  be  pursued  with  profit  at  this 
stage.  Cf.  the  author's  Differential  and  Integral  Calculus,  Appendix, 
p.  417. 


LOGARITHMS  AND   EXPONENTIALS  153 

or,  on  dropping  the  subscript : 

(6)  Dx\ogax  =  1-^^- 

x 

Thus  if  the  usual  base,  a  =  10,  be  taken,  the  formula 
becomes : 

/irk  n  i  .4343... 

x 

Discussion  of  the  Result.  We  have  met  a  similar  situation 
before,  in  the  differentiation  of  the  sine.  There,  if  angles  be 
measured  in  degrees,  the  fundamental  formula  reads  : 

DT  sin  x  =  -^—  cos  x. 
180 

In  order  to  get  rid  of  this  inconvenient  multiplier,  we 
changed  the  unit  of  angle  from  the  degree  to  the  radian,  and 
then  the  formula  became  : 

Dx  sin  x  =  cos  x. 

In  the  present  case,  it  is  possible  to  do  a  similar  thing.  The 
base,  a,  is  wholly  in  our  control,  to  choose  as  we  like.  Now, 
for  any  base,  the  logarithm  of  the  base  is  unity,  §  1,  (2) : 

log,  a  =  1. 

If,  then,  we  choose  as  our  base  the  number  e : 

a  =  e  =  2.71828  ... 

the  multiplier  becomes 

(8)  loga  e  =  loge  e  =  1. 

For  this  reason,  e  is  taken  as  the  base  of  the  logarithms 
used  in  the  Calculus.*  These  are  called  natural  logarithms. 
They  are  also  called  hyperbolic,  or  Naperian  logarithms,  —  the 
latter  name  after  Napier,  the   inventor   of  logarithms.     But 

*  The  notation  e  for  this  number  is  due  to  Euler,  1728. 


154  CALCULUS 

Napier*  was  the  very- man  who  introduced  denary  logarithms 
into  mathematics,  and  so  the  use  of  his  name  in  connection 
with  natural  logarithms  is  misleading. 

Since  natural  logarithms  are  always  meant  in  the  formulas 
of  the  calculus,  unless  the  contrary  is  explicitly  stated,  it  is 
customary  to  drop  the  index  e  from  the  notation  loge  x  and 
to  write 

(9)  y  =  log  x,         if         ey  =  x. 

The  identities  (G)  and  (7)  of  §  1  now  take  on  the  form : 

(10)  eXo*x  =  x, 

(11)  log  ez  =  x. 

The  formula  of  differentiation  becomes  : 


(12)                                     Bx\ogx  =  -. 

X 

In  differential  form  it  reads  : 

(13)                                      flogz=i, 
dx              x 

(14)                                         dlogcc  =  — • 

X 

Example.     Differentiate  the  function 

u  =  log  sin  x. 

Let                                      y  =  sin  x. 

Then                                       u  =  log  y, 

du  =  fZlog  y  =  -+ ,                 dy  = 

y 

cos  xdx, 

and 


,        cos  a;  da;  , 

du  =  — =  cot  xdx. 


sin  ./• 
*  Napier  was  a  Scotchman,  and  his  discovery  was  published  in  1614. 


LOGARITHMS  AND   EXPONENTIALS 
Hence  d  log  sin  x  =  cot  xdx, 


155 


—  log  sin  x  =  cot  x. 
dx 


EXERCISES 

Differentiate  the  following 

functions. 

1.    u  =  log  COS  X. 

du 

—  =  —  tan  x. 

dx 

2.    u  =  log  tan  x. 

—  =  cot  x  -f  tan  x 
dx 

3.    u  =  log  cot  X. 

du        -  2 
dx      sm2# 

4.    u  =  log  sec  x. 

5.    u  =  log  CSC  X. 

6.    u  =  log-^ — 

1  —  X 

du  _  1          1 
cte      x      1  —  x 

i      a  +  x 

7.    ii=  log— !— • 

a  —  .r 

du         2  a 

dx     a2  —  xl 
du          x 

8.    w  =  log  Va2  +  x2. 

dx     a2  +  x2 

9.    m  =  log  (1  —  cos  x). 

du          ,x 
—  =  cot  — 
dx            2 

10.    u  =  log  (1  +  cos  x). 

du                x 

—  =  —  tan  -  • 

dx               2 

3.    The  Limit  lira  (1  +  () '.     Since  this  limit  is  fundamental 

in  the  differentiation  of  the  logarithm,  a  detailed  discussion 
of  it  is  essential  to  completeness.     Let  us  set 

(i)  s=(i  +  0« 

and  compute  the  value  of  s  for  values  of  t  near  0.     Suppose 

t  =  .1.     Then 

s  =  (l.l)io, 


156  CALCULUS 

and  this  number  is  found  by  the  usual  processes  with  loga- 
rithms to  be  2.59. 

Further  pairs  of  corresponding  values  (t,  s)  are  found  in  a 
similar  manner.  In  particular,  the  student  can  verify  the 
correctness  of  the  following  table  of  values  :* 

t     I     -0.1  -.01  -.001     .     .     .      +.001  +.01  +0.1 

s    I         2.87  2.73  2.72       .     .     .       2.72  2.70  2.59 

The  foregoing  table  indicates  strongly  that,  when  t  ap- 
proaches the  limit  0  from  either  side,  the  variable  s  is 
approaching  a  limit  whose  value,  to  three  significant  figures,  is 
2. 71'.  This  is  in  fact  the  ease.t  The  exact  value  of  the  limit 
is  denoted  by  the  letter  e  : 

i 
(2)  lim(l  +  0'  =  e  =  2.71828 -. 

4.  The  Compound  Interest  Law.  The  limit  (2)  of  §  3  pre- 
sents itself  in  a  variety  of  problems,  typical  for  which  is  that 
of  rinding  how  much  interest  a  given  sum  of  money  would 
bear  if  the  interest  were  compounded  continuously,  so  that 
there  is  no  loss  whatever.  For  example.  $  1000,  put  at  in- 
terest at  6%,  amounts  in  a  year  to  $1060,  if  the  interest  is 
not  compounded  at  all.  If  it  is  compounded  every  six  months, 
we  have  .  ., 

$  1000(1  +  ~ 

as  the  amount  at  the  end  of  the  first    six   months,  and    this 
must  be  multiplied  by  [1+^—  )  to  yield  the  amount  at  the 

end  of  the  second  six  months,  the  final  amount  thus  being 

nooo/i  +  '-yY. 

*  To  compute  the  middle  entries  in  this  table  a  six-place  table  of 
logarithms  is  needed. 

t  For  a  rigorous  proof  cf .  the  author's  Differential  and  Integral  Cal- 
culus, p.  79. 


LOGARITHMS   AND    EXPONENTIALS  157 

It  is  readily  seen  that  if  the  interest  is  compounded  n  times 
in  a  year,  the  principal  and  interest  at  the  end  of  the  year  will 
amount  to  .         nfiN 

1000(1+  —  ] 

dollars,  and  we  wish  to  find  the  limit  of  this  expression  when 
n  =  oo.     To  do  so,  write  it  in  the  form  : 

and  set  t  —  ' — .     The  bracket  thus  becomes 
n 

and  its  limit  is  e.     Hence  the  desired  result  is 
1000  e06=  1061.84.* 


EXERCISE 

If  $  1000  is  put  at  interest  at  4  <f0,  compare  the  amounts  of 
principal  and  interest  at  the  end  of  10  years,  (a)  when  the 
interest  is  compounded  semiannually,  and  (6)  when  it  is  com- 
pounded continuously.  Ayes.    A  difference  of  $  5.88. 

5.    Differentiation  of  e*.     Before  beginning  this  paragraph 
the  student  will  turn  to  Chap.  VIII  and  study  carefully  §  1. 
Since 

(1)  y  =  ex  and  x  =  log  y 

are  equivalent  equations,  the  former  function  can  be  differen- 
tiated by  taking  the  differential  of  each  side  of  the  latter 
equation: 

dx  =  a  log  y  =  -^  • 

y 

*  The  actual  computation  here  is  expeditiously  done  by  means  of 
series  ;  see  the  chapter  on  Taylor's  Theorem. 


158 
Hence 
or 
(2) 

(3) 


Fig.  51 
The  function 

(4)  y  =  a' 

could  be  differentiated  in  a  similar  manner.  It  is,  however, 
simpler  to  take  the  logarithm  of  each  side  of  (4)  and  then  dif- 
ferentiate the  new  equation: 

log  y  =  log  a1  =  a  log  a, 


(5) 


d  log  y  =  -^-z=dx  log  a. 

y 

dax  =  ax  \ogadx. 


Differentiation  ofx".     It  is  now  possible  to  complete  the  dif- 
ferentiation of  this  function  for  the  case  that  n  is  irrational. 


LOGARITHMS   AND    EXPONENTIALS 

Since  by  §  2, 

(10), 

x  =  elosx, 

it  follows  that 

xn  _  gnlcgx^ 

and  hence 

dx"  =  de"logx 

„  IO„  ,  n  dx 

X 

Thus  finally, 

_    nndx 

X 

(6) 

dxn  =  nxn~ldx, 

159 


no  matter  what  value  n  may  have,  provided  merely  that  n  is 
a  constant. 

Differentiation  of  f(x)'b(x\     Let  it  be  required,  for  example, 
to  differentiate  the  function 

y  =  xx. 

Here,  both  base  and  exponent  are  variable.     Begin  by  tak- 
ing the  logarithm  of  each  side  of  the  equation : 


Hence 


and  so,  finally, 
or 


log  y  =  log  xx  =  x  log  a;. 
d  log  y  =  d(x  log  a;), 

-^  =  (1  +  log  x)  dx, 

II 

dy  =  y  (1  -f  log  a;)  dx 
d  xT  =  xx  (1  +  log  x)  dx. 


The  general  case,  ,,  _  f(xy<t>z) 

can  be  treated  in  a  similar  manner. 


160 


CALCULUS 


6.    Graph  of  the  Function  xn.     For  positive  values  of  n  the 
curves 

y  =  xn 


y=i 


Fig.  52 


lie  as  indicated  in  the  figure.  When  n  =  1,  we  have  the  ray 
from  the  origin,  which  bisects  the  angle  between  the  positive 
axes  of  x  and  y. 


LOGARITHMS   AND    EXPONENTIALS  161 

When  n  >  1,  the  curve  is  always  concave  upward ;  when 
n  <  1,  it  is  concave  downward. 

All  the  curves  start  at  the  origin  and  pass  through  the 
point  (1,  1). 

For  values  of  x  >  1,  the  larger  n,  the  higher  the  curve  lies. 
For  values  of  x  <  1,  the  reverse  is  the  case. 

Let  x  have  any  fixed  value  greater  than  unity  :   x  =  x'  >  1. 

Consider  the  ordinate  .„ 

y  =  x'  . 

As  n  increases,  x'n  increases  continuously.  This  property  is 
the  basis  of  the  property  of  logarithms  included  in  the  word 
continuous. 

For  proofs  of  the  foregoing  statements  cf.  the  author's 
Differential  and  Integral  Calculus,  p.  27  and  Appendix,  p.  117. 

7.  The  Formulas  of  Differentiation  to  Date.  The  student 
will  now  bring  his  card  of  formulas  up  to  date  by  supple- 
menting it  so  that  it  will  read  as  follows  : 

General  Formulas  of  Differentiation 
I.  dcu  =  cdu. 

II.  d(u-\-  v)  =  du  -f  dv. 

III.  d(uv)=  udv  +  vdu. 

TTT  ,AA      vdu  — udv 

IV.  d(-]= 2 

\V  J  V2 

Special  Formulas  of  Differentiation 

1)  dc  =  0. 

2)  d  xn  =  nxn~l  dx. 

3)  d  sin  x  =  cos  x  dx. 

4)  d  cos  x  =  —  sin  x  dx. 

5)  d  tan  x  =  sec2  x  dx. 


162  CALCULUS 

6)  d  cot  x  =  —  esc2  x  dx. 

7)  d\ogx  =  ^. 

X 

8)  dex  =  exdx. 

9)  dax  =  a1  log  a  '/.<•. 

To  obtain  facility  in  the  use  of  the  new  results  it  is  desirable 
that  the  student  work  a  good  number  of  simple  exercises. 

Example  1.     To  differentiate  the  function 

u  =  e"x. 

Let  y  =  ax. 

Then  ?<  =  e"}    . 

cfo,  =  dev  =  ev  r7?/  =  ^"(arfa;). 
Hence 


deaI 

ax  i                              d     „ 
=  ae"x  «x             or             —  e"x  = 

dx 

=  ae'' 

Example  2. 
show  that  * 

If 

ii  =  A  cos  (wi  +  y), 

**  +  n*tt  =  0. 

(ft2 

To  do  this,  compute  first  — .     The  computation  is  readily 

dt 

effected  by  taking  the  differential  of  each  side  of  the  given 

equation : 

du  =  Ad  cos  (nt  +  y) 


=  A[-  sin (nt  +y)d (nt  +  y)] 

=  —  An  sin  (nt  +  y)  eft, 

*  Such  an  equation  as  the  following  is  railed  a  differential  equation, 
and  any  function  which,  when  substituted  for  u,  satisfies  the  equation  is 
called  a  solution. 


,    LOGARITHMS   AND    EXPONENTIALS  163 


—  =  —  An  sin  Cut  +  7). 
dt  v  /; 


\ 


dhi 
Next,  compute  — .     Since 

fdu\ 
d?u_d(du\  =  (XdtJj 
df^dtydt)         dt 

we  take  the  differential  of  each  side  of  the  equation  for  — : 

1  dt' 

df—  \=-  Andsm(nt+y) 

=  —  An  [cos  (nt  -f  y)d(nt  +  y)] 
=  —  An2  cos  (nt  +  y)  dt. 
Hence,  on  dividing  through  by  dt,  we  have  : 

—  =  —  An2  cos  (nt  +  y). 
dP  v         Y! 


If  now  we  multiply  the  given  value  of  u  by  n2  and  add  the 
oduct  to  the  value  just  obtainec 
tically  0,  i.e.  0  for  all  values  of  t : 


product  to  the  value  just  obtained  for  — -,  the  result  is  iden 


d-n 
df1 

+ 

nhi  =  0, 

EXERCISES 

Differentiate  the  following 

functions. 

1.   u  =  e~x\ 

—  =  -2xe~z\ 
dx 

2.    u  =  eslux. 

—  =eaiaxcosx. 
dx 

3      11.  =  ft*  -I-  p-^2 

*H  =  2  (e2*  -  e~ 

q.  e. d. 


dx 


164  CALCULUS 

4.  a  =  10*.  — =  (2.30259  ...)10'. 

5.  a  =  .i-'o  10*.  —  =  x9 10* (10  +  2.30259a;). 

6.  ?t  =  log  (see  x  -f-  tan  x).  —  =  sec  a*. 

t/.c 

7.  it  =  oj2 logcc.  —  =  a;  (1  +  2  log  a). 

dx 

8.  m  =  oj3  log  (a  —  a:).  9.    u  =  e~*log(2x-\-3). 

10.    u  =  e~at  cos  (nJ  —  y).  11.    «  =  e~K'(^l  cos  nt-{-Bsh\  nt). 

a;  log  a;      ,      ,     .   1A         dw        log  a; 

12.    m  =  — 2- log  (a; +  1)-        -t-=; — ^rV 

x  +  1  (te      (a;  +  1)-' 

du  1 


13.    a  =  log  (aj  +  Va;2  —  a2). 


14.    «  =  log  (a;  +  Va2  +  x2). 


15.    « .  =  log  (ex  +  e_I). 
17.    u  —  log  tan-  • 


18.    w  =  log  tan  [  -  +  - 


19.    a  =  eot  f 


20.    M  =  tanf---\ 
\2      4; 


(/.<■ 

V.r2  -  o2 

du 

1 

dx 

Va2  +  a-! 

16.     M  = 

sin  ./•  +  eos  x 

daj 

=  CSC  X. 

<lu 

-  sec  0. 

du 

1 

dx 

1  —  sin  .<■• 

du 

1 

dx      1  +  sin  x 


sin  a: 


21.    «  =  log  Vl  +  sin  6.  22.    u  =  log 

x 


23.    u  =  log  Vl  —  cos  x.  24.    u  =  yV. 

25.    «=(10l+<)2.  26.    u=J/ar*. 

27.    M=f-^-Y.  28.    U—y/W. 


LOGARITHMS   AND    EXPONENTIALS  1G5 

29.  u  =  afmz.  —  =  af1111"1  (sin  x  +  x  cos  x  log  a;). 

dx  ' 

30.  «  =  (sin  .*-)COSl.      —  =  (sin  x)"os  r_1  (cos2  a-  —  sin2  x  log  sin  a). 

i 

31.  v=xx.  32.  w=  (cos  xyinx.  .  33.    M  =  (tan  x)x. 

34.    M  =  (loga?2)*.      35.   M  =  (l  +  a)°.  36.    «  =  (x2)21. 

37.    If  w  =  A  i-os,  nt  +  .B  sin  ut,  show  that 

+  n?u  =  0. 

dt1 


38.    If  u  =  Ce_K'  cos  (V«2  —  K2t  -4-  y),  show  that 

d2w  .  o    (/«  .     .>        A 
dt2  dt 


CHAPTER    VII 
APPLICATIONS 

1.  The  Problem  of  Numerical  Computation.  It  often 
happens  in  practice  that  we  wish  to  solve  a  numerical  equa- 
tion in  one  unknown  quantity,  or  a  pair  of  simultaneous 
equations  in  two  unknowns,  to  which  the  standard  methods 
with  which  we  are  familiar  do  not  apply ;  for  example, 

cos  x  =  x, 

j  2  cot  6  +  2  =  cot  0, 
[2cos0  +  cos<£  =  2. 

Such  equations  usually  come  to  us  from  physical  problems, 
and  the  solution  is  required  only  to  a  limited  degree  of 
accuracy,  —  say,  to  two,  three,  or  possibly  four  significant  fig- 
ures. Any  method,  therefore,  which  yields  an  approximate 
solution  correct  to  the  prescribed  degree  of  accuracy  furnishes 
a  solution  of  the  problem. 

In  particular,  the  problem  of  the  determination  of  the 
error  in  the  result  due  to  errors  in  the  observations  comes 
under  this  head. 

2.  Solution  of  Equations.     Known  Graphs. 
Example  1.     Let  it  be  required  to  solve  the  equation 

1)  cos  x  =  x. 

We  can  evidently  replace  this  problem  by  the  following : 
To  find  the  abscissa  of  the  point  of  intersection  of  the  curves 

2)  y  =  cos  x,  y  =  x. 


APPLICATIONS  167 

The  first  of  these  curves  we  have  plotted  accurately  to  scale. 
The  second  is  the  right  line  through  the  origin,  which  bisects 
the  angle  between  the  positive  coordinate  axes.  It  is,  there- 
fore, sufficient  to  lay  down  a  ruler  on  the  graph  of  the  former 
curve,  so  that  its  edge  lies  along  the  right  line  in  question, 
and  observe  where  this  line  cuts  the  curve.     The  result  lies 

between  _  ,  _ 

x  =  .7         and         x  =  .8, 

and  may  fairly  be  taken  as  x  =  .75.  It  is  understood,  as 
usual  in  approximate  values,  that  the  last  figure  tabulated  does 
'  not  claim  complete  accuracy ;  but  we  are  entitled  to  a  some- 
what better  result  than  would  be  given  by  the  first  figure 
alone. 

Example  2.     To  solve  the  equation 

3)  xs  +  2a:-2  =  0. 

Suppose  we  have  plotted  the  curve 

4)  y  =  x3 

accurately  from  a  table  of  cubes.  Then  the  problem  can  con- 
veniently be  formulated  as  follows  : 

To  find  the  abscissa  of  the  point  of  intersection  of  the  curves 

5)  y  =  x*  and  y  =  2  —  2x. 
The  details  are  left  to  the  student. 

Example  3.     To  find  the  positive  root  of  the  equation 

6)  e-i*  +  2.92a;  =  2.14. 

Here,  we  can  connect  up  with  the  graph  of  the  function  ex 
by  making  a  simple  transformation.     Let 

7)  o?'=— -la;;  x  =  -2x'. 
The  equation  then  becomes 

8)  e*'-5.84x'  =  2.14, 


168  CALCULUS 

and  we  seek  to  determine  the  abscissa  of  that  point  of  inter- 
section of  the  curves  (for  simplicity,  we  drop  the  accent) 

9)  y  =  ez  and  y  =  5.84a?  +  2.14 

which  lies  to  the  left  of  the  origin.  The  second  place  of 
decimals  in  the  coefficients  is  not  to  be  taken  too  seriously  ;  we 
make  as  accurate  a  drawing  as  the  graph  and  a  well-sharpened 
pencil  permit.  Having  thus  determined  the  negative  x'  from 
the  graphs  of  9),  we  find  the  desired  positive  x  by  substituting 
this  value  in  equations  7).  The  execution  of  the  details  is 
left  to  the  student. 

Example  4.     Solve  the  equation 

ex  =  tan  x,         0  <  x  <  y  ■ 

Lt 

If  one  of  the  curves 

y  =  e1  or      y  =  tan  x 

were  plotted  on  transparent  paper,  or  celluloid,  it  could  be 
laid  down  on  the  other  with  the  axes  coinciding  and  the  inter- 
section read  off.  The  same  result  can  be  attained  by  holding 
the  actual  graphs  up  in  front  of  a  bright  light. 

In  cases  as  simple  as  this,  however,  free-hand  graphs  will 
often  yield  a  good  first  approximation,  and  further  approxima- 
tions can  be  secured  by  the  numerical  methods  of  the  later 
paragraphs. 

EXERCISES* 

1.  Solve  the  equation 

cos  x  =  2x. 

2.  Find  the  root  of  the  equation 

3  sin  x  =  2x 
which  lies  between  0  and  jr. 

*  In  solving  these  exercises  only  so  great  accuracy  is  expected  as  can 
be  attained  from  well-drawn  graphs  of  the  standard  curves.  It  will  be 
shown  in  later  paragraphs  how  the  solutions  can  be  improved  analytically 
and  carried  to  any  desired  degree  of  accuracy. 


APPLICATIONS  169 

3.  Solve:  x  4-  tan  x  =  1,  0  <  a;  <  -. 

Li 

4.  Solve:  3  cos  cc  —  5#  =  6,         —  -  <  a;  <  0. 

5.  Find  the  root  of  the  equation 

log  x1  -f  2  =  a; 
which  lies  between  0  and  1. 

6.  Solve  :  sin  2x  =  x. 

7.  Find  all  the  roots  of  the  equation 

12a,-3  +  4a:  +  3  =  0. 

8.  The  same  for      „  ,  .       . 

ox6  —  5  x  —  1  =  0. 

9.  The  same  for 

ar  —  a  —  1  =  0. 

Solve  the  following  equations  : 

10.  cos3  $  4-  .47  cos  0  -  1.23  =  0,      0  <  6  <  90°. 


11.    sin  a*  =  V 1  —  x2.  12.    a,v-  4-  cos2  x  =  4. 

13.  Show  that  the  equation 

tan  x  =  x 

has  an  infinite  number  of  roots.     These  can  be  written  in  the 

form 

xn  =  -Mr  4-  e„, 

where  e„  is  numerically  small  when  n  is  numerically  large. 

14.  Find  the  largest  value  of  P  for  which  the  equation 

cos  a;  4-  Px  =  1 
admits  a  solution  in  the  interval   0  <  x  <  ir, 

15.  Find  the  point  of  the  parabola 

2y  =  x" 
which  is  nearest  to  the  point  (2,  0). 


170  CALCULUS 

16.    Find  the  radius  of  the  circle  whose  center  is  at  (0,  2) 
and  which  is  tangent  to  the  parabola 

y2  =  x. 
3.    Interpolation.     <  ionsider  the  equation 

I)  /(*)=<>. 

Suppose  a  root  has  been  located  with  some  deg 
More  precisely,  suppose  that 

ffa)  and  /(. 

are  of  opposite  signs.     If  the  function  f(x)  is  continuous  in 
the  interval  a^  ^  x  <  x2  and  if  its  derivative  is  always 
tive  (or  always  negative)  in  this  interval,  then  the  functi* 
always   increasing   (or  always  decreasing)  and  so  must  have 
just  one  root  between  xx  and  x2. 

The  root  can  be  found  approximately  as  follows.     Cons 
the  graph  of  the  function 

2)  j,  =/(*). 

Let  yx  =f(x1),  =f(&), 

and  draw  the   chord    through   the   poin  i   and    (x 

The  point  in  which  this  chord  cuts  the  axis  of  x  will  obvi 
yield  a  further  approximation  to  thi 
sought.     Denote  this  last  value  by  X. 
The  equation  of  the  chord  is 

3)  •''  =  ■--  =y  —  yi. 

■'■■  -  •'':        /'.  -  Vl 

On  setting  y  =  0  and  solving  for  x,  we 
have,  as  the  value  of  X.  the  following  : 

4)  X=x1-°^^ly1, 

y-i  -  ui 

or 

5)  X=x, ^— ^ — ff-y 


APPLICAT1  171 

-  -zplained  v  d  in  detail  loped,  in 

nnula  i  nina- 

proximation,  X.     In  pi 
ight  lin— 
-cale  and  read  off  from  the  fig"i 

-:der  equation  .  .    _ 

-  .     -2  = 
The  curve  in  question  is  here 

7)  =  -         - 

.aphical  solution       §  2 
z=  .7 

=      =  s  found  to  h:  line 

—    .  =  *  —       - 

rough  the  points 

=    '    -  .  and  ~         ::. 

■  -  .:    .i:_  -   257 

=  on  and  solving  for  ; 

X=.7  +     -      =.7 

see  about  ho~  rproximation  is,  eom- 

orresponding  value  o: 

>>3. 

jf  decimals,  z  =  " 

*  I"  ■  ■ 

^Heated  above  in  ti  He  should  take  10  cm.  to 

rv:;--  :.-  :i-  . -.  r:    :-.     :  .-..-i    1    ::.:_  .:.  =    7  to  z-:  =    - 


172  CALCULUS 

It  is  possible  to  apply  the  method  again,  taking  now 
(xb  ?/l)  =  (.7693,  -.0063) 

and  (#2,  y2)  as  before.  We  leave  this  as  an  exercise  to  the 
student.  He  should  make  both  the  graphical  determination 
with  an  enlarged  scale  and  the  analytic  determination  of 
formula  4). 

The  Method ;  Not,  the  Formula.  The  student  may  be 
tempted  to  use  the  formula  4)  or  5),  rather  than  to  go  back 
to  the  method  by  which  it  was  derived.  This  would  be  un- 
fortunate, for  the  formula  is  not  easily  remembered,  whereas 
the  method,  once  appreciated,  can  never  be  forgotten.  If  the 
student  finds  himself  in  a  lumber  camp  with  nothing  but  the 
ordinary  tables  at  hand,  he  may  solve  his  equation  if  he  has 
once  laid  hold  of  the  method.  It  is  true  that  the  best  way  is 
for  him  to  treat  first  the  literal  case  and  deduce  the  formula. 
But  this  he  may  not  be  able  to  do  if  he  has  relied  on  the 
formula  in  the  book. 

EXERCISES 

Apply  the  method  to  a  good  number  of  the  problems  at  the 
end  of  §  2. 

4.  Newton's  Method.  Suppose  again  that  it  is  a  question  of 
solving  the  equation 

1)  /(*)=»  0, 

and  suppose  we  have  already  succeeded  in  rinding  a  fairly 
good  approximation,  x  =  xv 

Consider  the  graph  of  the  function 

2)  y  =/(»). 

Compute  yt  =f(xi).     To  improve  the  approximation,  draw  the  . 
tangent  at  the  point  (xu  t/j).     Its  equation  is  : 

3>  »-*-(2m(— *  I 


APPLICATIONS  173 

Evidently,  this  line  will  cut  the  axis  of  x  at  a  point  very  near 
the  point  in  which  the  curve  2)  cuts  this  axis.  If,  then,  we 
set  y  =  0  in  3)  and  solve  for  x,  we  shall  ob-  y 
tain  a  second  approximation  to  the  root  of 
1)  which  we  seek.  The  value  of  this  root 
will  be 


4)  X  =  xx- 


^7V 
2/i  Fig.  54 


Example  1.  Let  us  apply  the  method  to  the  Example 
studied  in  §  3.  In  order,  however,  to  have  simpler  numbers 
to  work  with,  take  xx  =  .77  and  compute  the  corresponding 
'/i ;  it  is  found  to  be  :     y^  =  —  .0035. 

(x1}  jfc)  =  (.77,  -.0035). 

We  must  next  compute  dy/dx  from  the  equation 

y  =  x3  +  2x—2; 

*y  =  $x*  +  2,  (QC\        =3.779. 

dx  \dxJx=.n 

On  substituting  these  values  in  3),  we  have : 

y  +  . 0035=  3.779  (x  -  . 77). 

[Sow  set  y  =  0  and  solve.     The  result  is  that  given  by  4)  : 

„„  .  .0035      „_no 

x  =  .  /  i  H =  .  ii  09. 

3.779 

We  have  tabulated  four  figures  in  the  result  because  this  is 
bout  the  degree  of  accuracy  that  seems  likely.  To  test  this 
>oint,  compute  y  for  the  value  of  x  which  has  been  found : 

y|, =.7709  =  -.0001. 

iince  the  slope  of  the  graph  is  greater  than  unity,  the  error 
q  x  is  less  than  one  unit  in  the  fourth  place.  It  is  easy  to 
erify  the  result  by  computing  y  for  the  next  larger  four-place 
alue  of  x :  ,  ,    nnAo 


174  CALCULUS 

Thus  we  have  a  complete  proof  that  the  root  lies  between 
.7709  and  .7710,  and  we  see  that  it  lies  about  one  quarter  of 
the  way  from  the  first  to  the  second  value. 

Example  2.     It  is  shown  that  the  equation  of  the  curve  in 
which  a  chain  hangs,  —  the  Catenary,  —  is 

5)  '-$(•* +*~*)  i 

where  a  is  a  constant.  The  length  of  the  arc,  measured  from 
the  vertex,  is 

6)  s=£(V-e" 


Let  it  be  required  to  compute  the  dip  in  a  chain  32  feet  long, 
its  ends  being  supported  at  the  same  level,  30  feet  apart. 

We  can  determine  the  dip  from  5)  if  Ave  know  a,  and  we 
can  get  the  value  of  a  from  6)  by  setting  s  =  16,  x  =  15  : 

/  is  i; 

Let  x  =  — . '    Then 
a 

f(x)  =  e*-  e~x  -  fix  =  0, 

and  we  wish  to  know  where  the  curve 

7)  y  =f(x)  =  ex  -  er-  -  ff a; 

crosses  the  axis  of  x. 

This  curve  starts  from  the  origin  and,  since 


!=/'(*)= «■+*--« 


TS 


is  negative  for  small  values  of  x,  the  curve  enters  the  fourth 
quadrant.     Moreover, 

cpL=eI-e-x>0,  x>0, 

ax2 


APPLICATIONS  175 

and  hence  the  graph  is  always  concave  upward.     Finally, 

/(l)  =  e-e-i-2^.  =  .217>0, 

and  so  the  equation  has  one  and  only  one  positive  root,  and 
this  root  lies  between  0  and  1. 

It  will  probably  be  better  to  locate  the  root  with  somewhat 
greater  accuracy  before  beginning  to  apply  the  above  method. 
Let  us  compute,  therefore,  /(^).  By  the  aid  of  Peirce's  Tables 
we  rind  : 

/(.5)  =  1.6487  -  .6065  -  1.0667  =  -  .0245  <  0. 

Comparing  these  two  values  of  the  function : 

/(.5)  =  -.02,  /(1)  =  .22, 

and  remembering  that  the  curve  is  concave  upward,  so  that 
the  root  is  somewhat  larger  than  the  value  obtained  by  direct 
interpolation  (this  value  corresponding  to  the  intersection  of 
the  chord  with  the  axis  of  x)  we  are  led  to  choose  as  our  first 
approximation  xl  =  .6  : 

/(.6)  =  1.8221  -  .5488  -  1.2800  =  -.0067, 

/'(.6)  =  1.8221  +  .5488  -  2.1333  =     .2376. 

Hence  the  value  of  the  next  approximation  is 

X  =  .6  -  ~  •0067  =  .6  +  -0282  =  .628. 
.2376 

To  get  the  next  approximation  we  compute 

/(.628)  =  1.8739  -  .5337  -  1.3397  =  .0005. 

[iHence  the  value  of  the  root  to  three  significant  figures  is  .628 
with  a  possible  error  of  a  unit  or  two  in  the  last  place,  and  the 
.value  of  a  we  set  out  to  compute  is,  therefore,  15/.628  =  23.9. 

Remark.  Newton's  method,  like  the  other  methods  of  this 
'chapter,  has  the  advantage  that  an  error  in  computing  the  new 
approximation  will  not  be  propagated  in  later  computations. 
Such  an  error  will  in  general  hinder  us,  because  we  are  not 


176  CALCULUS 

likely  to  get  so  good  an  approximation.  But  the  one  test  for 
the  accuracy  of  the  approximation  is  the  accurate  computa- 
tion of  the  corresponding  y,  and  if  this  is  done  right,  we  see 
precisely  how  close  we  are  to  the  desired  root. 

The  function  f(x)  is  usually  simple,  and  it  is  easy  to  see 
whether  the  curve  is  concave  upward  or  concave  downward 
near  the  point  where  it  crosses  the  axis.  "We  thus  have  a 
means  of  improving  the  approximation  at  the  same  time  that 
we  simplify  the  new  value  of  x.  For,  if  the  curve  lies  to  the 
right  of  its  chord,  the  approximation  by  interpolation  will  be 
too  small ;  and  if  the  curve  lies  to  the  right  of  its  tangent  be- 
tween the  point  of  tangency  and  the  axis  of  x,  the  approxima- 
tion given  by  Newton's  method  will  also  be  too  small. 

Comparison  of  the  Two  Methods.  When  looked  at  from  their 
geometric  side  the  two  methods  appear  much  alike,  the  first 
seeming  somewhat  simpler,  since  it  does  not  involve  the  use 
of  derivatives.  Why  bother,  then,  with  Newton's  method  ? 
It  is  not  a  theoretical  question,  but  purely  one  of  convenience 
in  carrying  out  the  numerical  work.  It  will  be  found  that,  as 
a  rule,  the  first  method  is  preferable  in  the  early  stages 
(usually,  merely  in  the  first  stage).  When,  however,  a  fairly 
good  approximation  has  been  reached,  the  numerical  work  in- ' 
volved  in  Newton's  method  is  generally  shorter  than  that 
required  by  interpolation. 

EXERCISES 

Apply  the  method  to  the  Exercises  of  §  2.     When,  however,  i 
bhe  approximation  given  by  the  graphical  method  of  §  1  is 
crude,  the  method  of  interpolation  may  be  used  to  improve  it. 

5.   Direct  Use  of  the  Tables. 

Example  1.  Let  us  recur  to  the  first  example  studied,  Ex.  1, 
§2: 

1)  COS  X  =  X. 


APPLICATIONS 


177 


The  graphical  solution  gave  x  =  .75.  Turn  now  to  a  table 
?f  natural  cosines  in  radian  measure,  preferably  Peirce's  Tables. 
A.s  we  run  down  the  table,  we  find  the  entries  : 


RADIANS 


.7389 
.7418 


COS    NAT 


.7392 
.7373 


Thus  x  is  seen  to  lie  between  .7389  and  .7418.  It  is  an  ex- 
jellent  exercise  for  the  student  to  work  out  the  interpolation 
tor  himself  before  we  take  it  up  at  the  end  of  the  paragraph, 
rhe  answer  is  :  x  =  .7391. 

Example  2.     Consider  the  equation 
I)  tan  x  =  e1, 

;he  desired  root  lying  between  0  and  tt/2. 

A  free-hand  drawing  of  the  graphs  of  the  functions 

y  =  tan  x,  y  =  ex 

hows  that  x  lies  between  1  and  1.5.  So  the  next  step  is  taken 
conveniently  by  opening  Peirce's  Tables  to  the  Trigonometric 
^unctions  and  Huntington's  to  the  Exponentials,  and  writing 
jrown  the  two  pairs  of  values  of  the  functions  which  came 
[.earest  together : 

x  tan  x  ex 


1.3 
1.4 


3.60 

5.80 


3.67 
4.06 


Thus  the  root  is  seen  to  lie  between  1.3  and  1.4. 

The  general  case  which  the  above  examples  are  intended  to 
lustrate  is  the  following :  —  To  solve  the  equation 

f(x)  =  <f>(x), 

here  f(x)   and   <£(#)   are   tabulated   functions,   or   functions 
;adily  computed. 


178 


CALCULUS 


When  the  solution  has  progressed  to  the  point  indicated 
by  the  examples,  the  next  step  can  be  taken  by  interpolation, 
or  by  Newton's  method,  as  will  now  be  explained. 

Interpolation.  When  two  values  of  the  independent  varia- 
ble near  together,  xx  and  x2,  have  been  found  such  that  /(.r)"is 
greater  than  <f>(x)  for  one  of  them  and  less  than  cf>(x)  for  the 
other,  the  best  approximation  to  take  next  is  the  one  given  by 
the  abscissa  of  the  point  of  intersection  of  the  chords  of  the 
graphs  of  the  functions, 

This  value,  X,  can  be  found  as  follows. 
Suppose  that 

/(Oi)  <  c^a-j)  and  /(a*)  >  <f>(x2).    ■ 

Introduce  the  following  notation  : 


4(x1)-f(x1)  =  ±u 

X2  —  x±  =  0, 


<P(x) 


V 

An 

Bt 

--y&£ 

Ay 

B2 

0 

r,         X 

C2 

Fig.  55 


f(x2)-<t>(x2)  =  &2, 

X  —  Xi  =  h. 

From  the  figure,  the  triangles 
A1CB1  and  A2CB2  are  similar, 
and 

A&  =  Aj,  A-,B2  =  A2. 

Their  altitudes,  when  C  is  taken 
as  the  vertex,  are  respectively  h 
and  8  —  h.     Hence 

Ax         A2 


3) 


On  solving  this  equation  for  h  we  find  : 

A, 


If/(xO  >  ^(xj)  and/(.r2)  <  <K-*'2)>  the  result  still  holds,  for 
Ax  and  A2  now  become  negative,  but  their  numerical  values 


APPLICATIONS  179 

correspond  to  the  lengths  of  the  sides   of   the   triangles   in 
question. 
It  is  easy  to  express  in  words  the  result  embodied  in  3). 

Rule.     In  order  to  see  what  fraction  of  8  =  x2  —  xx  must  be 
added  to  xt  in  order  to  give  X,  form  the  differences 

<f>(xi)-f(xi),  /(&)  -  <K«2)- 

Tlien  the  fraction  is  the  quotient  of  the  first  of  these  differences  by 
},heir  sum. 

In  practice,  an  accurately  drawn  figure  on  a  large  scale  will 
Dften  afford  a  quicker  and  sufficiently  accurate  solution. 


Example.     Returning  to  Ex.  1  above,  we  have  : 
f(x)  =  cos  x,  <f>(x)  =  x ; 

8  =  x2-  Xi  =  .0029,         xy  =  .7389,         x2  =  .7418. 
$(»0  -  f(xx)  =  -  .0004  ;  f(x2)  -  4>(x2)  =  -  .0045. 

•0004 .0029  =  ^116  =.0002. 


.0049  49 

3ence  the  value  of  the  new  approximation  is 
X  =  .7389  +  .0002  =  .7391. 

The  student  will  have  no  difficulty  in  completing  Ex.  2  above 
a  a  similar  manner.  It  turns  out  that  the  correction  is  here 
ess  than  one  tenth  of  8,  and  hence  it  does  not  influence  the 
econd  place  of  decimals  :     x'  —  1.30. 

Newton' 's  Method.  If  a  higher  degree  of  accuracy  is  desired, 
t  is  well  now  to  apply  Newton's  method  to  the  function 

F{x)  =  f{x)-^{x). 

In  the  case  of  Ex.  1  above  it  is  pretty  clear  that  we  already 
ave  four-place  accuracy,  and  the  computation  of  F(x)  for  the 
alue  X  =  .7391  would  only  verify  the  result.  This  is  as  far 
3  we  can  go  with  four-place  tables.     If  we  needed    greater 


180  CALCULUS 

accuracy,  we  should  use  Newton's  method  and  five  or  six-place 
tables. 

Example  2  has  been  carried  only  to  two-place  accuracy,  or 
three  significant  figures.  We  can  obtain  two  further  figures 
with  the  tables  at  our  disposal. 

y  =  F(x)  =  tan  x  —  ez. 

?/,  =  F(1.30)=  3.602  -  3.669  =  -  .067. 


-2-  =  sec-  x  —  t'r, 

dx  clx 


=  13.97  -  3.67  =  10.30 

i     L.30 

.067* 


X  =  1.30  +  —  =  1.3067. 
10.3 

To  test  this  result,  however,  would  require  five-place  tables. 


EXERCISES 
Solve  the  following  equations  : 
1.    cot  x  =  x,         0  <  x  <  tv.  2.    ex  +  log  x  =  1. 

3.  The  hyperbolic  sine  (sh  x  or  sinh  x)  and  cosine  (ch  x  or 
cosh  x)  are  defined  as  follows  : 

,  ex  —  e'x  n  e*  +  e~* 

sh  x  =  -  ,  ch  a;  =       — 

2      '  2 

and  are  tabulated  in  Peirce's  Tables,  pp.  120-123.  By  means 
of  these,  reduce  the  treatment  of  Ex.  2,  §  4,  to  the  methods  of 
the  present  paragraph. 

6.  Successive  Approximations.  We  come  now  to  one  of  the 
most  important  of  all  the  methods  of  numerical  computation. 
In  physics  it  is  known  as  the  method  of  Trial  and  Error :  in 
mathematics  it  goes  under  the  name  of  the  method  of  Succes- 
sive Approximations. 

The   problem   is   that   of   solving  a   pair   of    simultaneous 
equations, 
1)  F(x,y)=0,  Hx,y)  =  0. 


APPLICATIONS 


181 


The  cases  which  arise  in  practice  are  characterized  in  general 
by  two  things  :  First,  there  is  only  one  solution  of  the  equa- 
tions which  interests  us,  and  the  physical  problem  enables  us 
to  make  a  fairly  good  guess  at  it  for  the  first  approximation. 
Secondly,  each  of  the  equations  1)  is  simple,  the  curve  can 
readily  be  plotted  in  character,  and  the  equation  can  be  solved 
with  ease  numerically  for  the  dependent  variable  when  a  nu- 
merical value  has  been  given  to  the  independent  variable.  But 
elimination  of  one  of  the  unknowns,  though  sometimes  possible, 
is  not  expedient,  since  the  resulting  equation  is  hard  to  solve. 
The  method  is  as  follows.  Plot  the  curves  1)  in  character 
with  sufficient  accuracy  to  determine  which  of  them  is  steeper 
(i.e.  has  the  numerically  larger  slope)  at  their  point  of  inter- 
section.    Let 

(7,:  F(x,y)=0        or        y=f(x) 

be  the  one  that  is  less  steep, 

Co:  &(x,y)=0         or         x  =  <f>(y), 


y 

Cj 

/(\ 

y-2 



/•   ' 

X 

0 

c,  x3 

y 

C>\ 

Vi 

r^^^i 

y3 

1     \ 

1 

1 

1 

1 
1 

1 

X 

0 

h 

Fig.  56  Fig.  57 

the  other.     Then,  making  the  best  guess  we  can  to  start  with, 
x  =  xu  compute  y1  from  the  equation  of  C, : 

yi  =/Oi), 

and  substitute  this  value  in  the  equation  of  C2,  thus  getting 
the  second  approximation : 

x-i  =  <f>(yi). 

Proceeding  with  x2  in  the  same  manner,  we  obtain  first  y2, 
then  x3,  and  so  on. 


182  CALCULUS 

The  successive  steps  of  the  process  are  shown  geometrically 
by  the  broken  lines  of  the  figures. 

The  success  of  the  method  depends  on  the  ease  with  which 
y  can  be  determined  when  x  is  given  in  the  case  of  Cu  while 
for  C2  x  must  be  easily  attainable  from  y.  If  the  curves  hap- 
pened to  have  slopes  numerically  equal  but  opposite  in  sign, 
the  process  would  converge  slowly  or  not  at  all.  But  in  this 
case  the  arithmetic  mean  of  x1  and  x2  will  obviously  give  a 
good  approximation. 

The  method  has  the  advantage  that  each  computation  is 
independent  of  its  predecessor.  An  error,  therefore,  while  it 
may  delay  the  computation,  will  not  vitiate  the  result. 

Example.  A  beam  1  ft.  thick  is  to  be  inserted  in  a  panel 
10  x  15  ft.  as  shown  in  the  figure.  How  long  must  the  beam 
be  made  ? 

We  have : 

f  sin  <f>  +  I  cos  <f>  =  15, 
|  cos  <f>  +  I  sin  <f>  =  10. 

Hence  cos2  <f>  —  sin-  <f>=  10  cos  <j>  —  15  sin  </>. 

Fig.  58        Now  an  expression  of  the  form 

a  cos  <f>  —  b  sin  <f> 

can  always  be  written  as 

Vet'  4-  b-[        a       cos  t£ sin  <f> )  =  Va2  +  b2  cos  (<£  +  a), 

Wa2+&2  Va2+62          / 

a  ■  h 

where         cos  a  = — ,  sm  a  = 


Va-  +  62  Va2  +  V- 

In  the  present  case,  then, 

cos  2  <f>  =  V325  cos  (<£  +  a), 

,  10  •  15 

where  cos  «  =  —    =,  sin  a  = 


V325  V325 

Thus  «  is  an  angle  of  the  first  quadrant  ami 
tan  a  =  f ,  a  =  56°  10'. 


APPLICATIONS  183 

Our  problem  may  be  formulated,  then,  as  follows :    To  find 
the  abscissa  of  the  point  of  intersection  of  the  curves  : 

y  =  cos  2  (f>,  y  =  V325  cos  (<£  +  a). 

We  know  from  the  figure  a  good   approximation  to  start 
with,  namely : 

tan  (f>  =  |,  </>  =  33°  44'. 

For  this  value  of  <f>  the  slopes  are  given  by  the  equations  :  * 
ISO  ,  dy_  =  _  2  sin  2^  =  _  2  sin  67°  28'  =  -  1.8, 

IT  Cl(j> 

—  •  ^  =  -  V325  sin  (<*>+«)  =  -  V325  =  -  18. 

7T  dfj>  V  ' 

Hence  we  have : 

Ci '.  y  =  cos  2  <f>  ; 

C2 :     v  =  V325  cos  (<£  +  a)         or         $  =  cos-1  — ^ a. 

V325 
Beginning  with  the  approximation     . 

fa  =  33°  44', 

we  compute  yx  =  cos  67°  28'  =  .3832. 

!  Passing  now  to  the   curve    C2,  we   compute   its    <£    when   its 

>  V  =  !h  ■ 

.3832  =  V325  cos  (fa  +  a),  fa  =  32°  31'. 

We  now  repeat  the  process,  beginning  with  fa  =  32°  31'  and 

y/2  =  cos65°  02' =  .4221, 

.4221  =  V325  cos  (fa  +  «),  fa  =  32°  23'. 

A  further  repetition  gives  fa  =  32°  22',  and  this  is  the  value 
of  the  root  we  set  out  to  determine. 

*  Since  the  degree  is  here  taken  as  the  unit  of  angle,  the  formulas  of 
lifferentiation  involve  the  factor  -ir/180  ;  cf.  Chap.  V,  §  2. 


184  CALCULUS 

EXERCISES 

1.  Solve  the  same  problem  for  a  beam  2  ft.  thick. 

2.  A  cord  1  ft.  long  has  one  end  fastened  at  a  point  0  2  ft. 
above  a  rough  table,  and  the  other  end  is 
tied  to  a  rod  2  ft.  long.  How  far  can  the 
rod  be  displaced  from  the  vertical  through 

-     O  and   still   remain  in  equilibrium  when 


released  ? 


The  equations  on  which  the  solution  depends  are : 
I  2  cot  6  +  -=  cot  (£, 

I  2  cos  6  +  cos  <£  =  2. 
If  the  coefficient  of  friction  p  =  },-,  find  the  value  of  <£. 

3.  A  heavy  ring  can  slide  on  a  smooth  vertical  rod.  To 
the  ring  is  fastened  a  weightless  cord  of  length  2a,  carrying  an 
equal  ring  knotted  at  its  middle  point  and  having  its  further 
end  made  fast  at  a  distance  a  from  the  rod.  Find  the  position 
of  equilibrium  of  the  system. 

4.  Solve  Example  2,  §  4,  by  the  method  of  successive  ap- 
proximations. 

7.   Arrangement  of  the  Numerical  Work  in  Tabular  Form. 

In  the  foregoing  paragraphs  we  have  laid  the  chief  stress  on 
setting  forth  the  great  ideas  which  underlie  these  powerful 
methods  of  numerical  computation.  There  are,  however,  cer- 
tain details  of  technique  which  are  important,  not  only  for 
ease  in  keeping  in  view  the  results  obtained,  but  also  for 
accuracy,  since  they  reduce  the  numerical  work  to  a  system. 
We  will  illustrate  what  we  mean  by  an  example. 

Example.  Let  it  be  required  to  find  all  the  values  of  x  be- 
tween 0°  and  360°  which  satisfy  the  equation 

sin  x  =  log10  (1  —  cos '  x). 


APPLICATIONS 


185 


A  free-hand  graph  of  each  of  the  functions 

1)  y  =  sin  x,  y  =  log10  (1  —  cos  x) 

shows  that  there  is  one  root  between 
0°  and  180°  and  a  second  between 
180°  and  3G0°.  But  these  roots 
cannot  be  located  with  any  great 
accuracy  in  this  manner.  It  is  nec- 
essary to  do  exact  table  work,  and 
to  keep  the  successive  results  in 
such  form  that  they  are  convenient 
for  later  reference. 

To  this  end  such  a  table  as  the  following  is  useful.* 
with  the  trial  value  x  =  150°. 


Fig.  60 


Begin 


X 

150° 

cosx 

1  —  COS  X 

logio  (1  —  cosx) 

-  .8660 

1.8660 

.2709 

sinx 

.5000 

Since  the  ordinate  of  the  sine  curve  is  larger  than  that  of  the 
logarithmic  curve,  it  is  clear  from  the  figure  that  x  is  too  small. 
Try  x  =  160°. 

Before  proceeding  further  let  us  ask  ourselves  whether  the 
above  scheme  is  the  simplest  for  the  example  in  hand.  For 
the  special  value  x  =  150°  we  know  cos  x  without  reference  to 
the  tables,  and  hence  one  entry  of  the  tables  was  sufficient. 
But  when  x  =  160°,  it  will  be  necessary  to  enter  the  tables  first 
for  cos  x,  a  second  time  for  log10  (1  —  cos  x),  and  still  a  third 
time  for  sin  x. 


*  Paper  ruled  in  small  squares  is  convenient  for  these  tables,  the  in- 
i  dividual 'digits  being  written  in  separate  squares. 


186 

Now, 


CALCULUS 


I  —  cos  x  =  2  sirt 


,# 


2' 


logic,  (1  -  cos  aj)  =  log10  sitf  -  +  logjo  2 
=  2  log  sin  *  +  -3010. 

Heuce  it  is  possible  to  get  along  with  only  two  entries  of  the 

tables  if  we  make  use  of  the  following  scheme. 


X 

160° 

164° 

163°  3' 

\* 

80° 

82° 

81°  32' 

log10  sili  \  x 

1.9934 

1.9958 

1.9952 

2  logio  sin  ?,  x 

1.9868 

1.9916 

1.9904 

+  .3010 

.2878 

.21»2(J 

.2914 

sin  (180 -x) 

.3420 

.2756 

.2910 

The  ordinate  of  the  sine  curve  is  still  in  excess,  but  only 
slightly  so.  Try  x  —  164°.  It  is  seen  that  the  curves  have 
now  crossed.  Moreover,  the  two  approximations  for  x  — 
namely,  160°  and  164°  —  are  so  near  together  that  we  can  with 
advantage  apply  the  method  of  interpolation  of  §  5.  We 
have 


<f>  (.f)  =  sin  x 

/(»)  =  log10 

(1 

—  COS  x)  ; 

«!  =  160°, 

x,  =  164°, 

8  =  4°; 

+  (x0  =  .3420, 

fix,)  =  .2878, 

Ai  =  .0542 ; 

<£(.«./)  =  .2750, 

/(a*)  =  .2926, 

A2  =  .0170; 

h  =  - 
A: 

Ai      8._-05424_ 
+  A2         .0712 

:  3. 

05. 

Thus    the    correction    is    seen   to   be   3.05°,    or    3°  3',    and 
the  new  approximation  is  : 

x  =  163°  3'. 


APPLICATIONS  187 

For  this  value  of  x  the  values  of  the  two  functions,  f(x)  and 
<f>  (x),  differ  by  a  quantity  which  is  comparable  with  the  error 
of  the  tables,  and  the  problem  is  solved. 

EXERCISES 

1.  Determine  the  other  root  in  the  above  problem. 

2.  Solve  the  equation  : 

cot  x  =  login  (1  +  sin  x),  0  <  x  <  90°. 

3.  Find  the  positive  root  of  the  equation 

erx  =  x3  —  x. 

Suggestion.  Tabulate  x,  x3  (from  a  table  of  cubes),  x3  —  x, 
and  e~x. 

8.  Algebraic  Equations.  By  an  algebraic  equation  is  meant 
an  equation  of  the  form 

1)  aaxn  +  a^-1  +  •••  +  a„  =  0,  a0  =£  0, 

where  n  denotes  a  positive  integer. 

If  the  coefficients  a0,  al5  •••  are  numerical,  the  roots  can  be 
approximated  to  by  the  method  of  interpolation  or  by  Newton's 
method.  In  either  case  it  becomes  necessary  to  compute  the 
value  of  the  polynomial 

f(x)  =  aux"  +  diX"'1  +  ••■  +  an 

for  several  values  of  x,  the  later  ones  of  which  will  be  at  least 
three-  or  four-place  numbers.  There  are  labor-saving  devices 
for  performing  these  computations,  to  which  we  now  turn. 

Numerical  Computation  of  Polynomials.  Let  a  cubic  poly- 
nomial, for  example,  be  given  : 

f(x)  =  ax3  +  bx2  +  cx  +  d, 

and  let  it  be  required  to  compute  f(x)  for  the  value  x  =  m. 
Write  down  the  following  scheme  : 

_a am  +  b am2  +  bm  +  c         f(m) 

am        am2  +  bm        amz  +  bm2  +  cm 


188  CALCULUS 

the  explanation  of  which  is  as  follows.  Begin  with  the  first 
coefficient,  a,  and  multiply  it  by  m  to  get  the  expression  am 
which  stands  below  the  line.  To  this  expression  add  the 
second  coefficient,  b,  to  get  the  second  expression  above  the 
line,  am  +  b.  Next,  multiply  this  expression  by  m  to  get 
the  expression  which  stands  below  it,  and  continue  the  process. 
The  last  entry  above  the  line  will  be  the  required  value, 

f(m)  =  am3  +  bml  +  cm  +  d. 

Exa  mple.     Let 

f(x)  =  7  Xs  —  6#-  +  3.r  —  7, 

and  let  it  be  required  to  compute  the  value  of  f(x)  for  x  =  .8. 
Here,  the  scheme  is  as  follows  : 

7  -.4  2.68  -4.856 

5.6         -  .32        2.144 

and  hence 

/(.8)  =  -  4.856. 

It  will  be  observed  that  the  process  requires  only  additions 
(or  subtractions)  and  multiplications.  The  former  can  be  per- 
formed mentally.  The  latter  are  executed  most  simply  by 
one  of  the  machines  now  in  general  use  with  computers. 
These  instruments,  combined  with  the  method  of  this  para- 
graph, have  rendered  Horner's  method  for  solving  numerical 
algebraic  equations  obsolete. 

EXERCISE 
Compute  the  value  of 

5.1.r,4  -  3.42a:-  +  1.432.r  +  .8543 
for  x  =  .1876. 

In  the  problems  which  arise  in  physics,  however,  it  is  not 
a  question  of  computing  all  the  roots  of  a  numerical  equation, 
about  which  nothing  is  known  beyond  the  coefficients.  Usu- 
ally, the  equation  is  a  cubic  or  biquadratic,  and  only  one  root 
is  required.     Moreover,  from  the  nature  of  the  problem,  a  close 


APPLICATIONS  189 

guess  at  the  value  of  this  root  can  be  made  at  the  outset. 
Then  the  methods  set  forth  in  this  paragraph  and  in  §§  2,  3 
lead  quickly  to  the  desired  result. 

EXERCISES 

Solve  the  following  equations,  being  given  that  there  is  one 
root,  and  only  one,  between  0°  and  90°  : 

1.  4  cos3  0  -  3  cos  0  =  .5283,  0°  <  0  <  90°. 

2.  sin3  0  -  .75  sin  0  =  .1278,  0°  <  0  <  90°. 

3.  Find  the  root  of  the  equation 

x4  +  2.6 : x3  -  5.2a;2  -  10.4.x  +  5.0  =  0 
which  lies  between  0  and  1. 

4.  Find  the  root  of  the  equation 

3  a;4-  12a3  +  12  a?2  -4  =  0 
which  lies  between  2  and  3. 

9.  Continuation.  Cubics  and  Biquadratics.  Aside  from  the 
special  problem  of  numerical  computation,  the  simpler  alge- 
braic equations  present  an  intrinsic  interest  which  should  not 
be  ignored. 

Transformations,     a)  Let  the  cubic  equation 

1)  f(x)  =  ax?  +  bx1  +  ex  +  d  =  0,  a  =£  0, 
be  given,  and  let  x  be  replaced  by  y,  where 

2)  y  =  x  —  h,  x  =  y  +  h. 

Then 

f(x)  =  a(y  +  hf  +  b{y  +  Kf  +c(y  +  h)  +  d  =  <f>(y) 

=  ay3  +  (3ah  +  b)y?+  -, 

where  the  later  coefficients  are  easily  written  down. 


190  CALCULUS 

If  y  =  p  is  a  root  of  the  equation 

3)  <Kv)  =  o, 

then  x  =  /3  +  /* 

will  be  a  root  of  equation  1).     For,  it  is  always  true  that 

/(*)-*  GO 

when  x  and  #  are  connected  by  the  relation  2). 

Here,  h  is  any  number  we  please.  In  particular,  h  can 
always  be  so  chosen  that  the  coefficient  of  the  second  term  of 

3)  will  drop  out.     It  is  sufficient  to  set 

4)  3ah  +  b  =  0,         or        fc  =  _A. 

3a 

Obviously,  the  same  method  can  be  used  to  transform  an 
algebraic  equation  of  any  degree  into  a  new  equation  whose 
second  term  is  lacking. 

EXERCISES 

Transform  the  following  equations  into  equations  in  which 
the  second  term  is  lacking. 

1.  x3  +  x1  —  x  +  1  =  0.  2.  3  x3  —  4  x2  +  2  =  0. 

3.  x4  +  .r3  —  x2  + 1  =  0.  4.  ox4  —  4x?  +  as2  -fa?  —  80=0. 

5.  3  x4  —  7  x3  +  a2  -  x  —  1  =  0.    6.  x6  +  x*5  +  x-  +  24-1  =  0. 

6)  Let  the  equation 

5)  f(x)  =  x*  +  pjc2  4  gas  4  r  =  0 
be  given,  and  let  x  be  replaced  by  y,  where 

6)  y=k'  x=ky. 
Then 

f(x)  =  fcy  +  feW  4  kg*/  +  r. 

Denoting  this  last  polynomial  by  <f>(y),  we  have 

/(«)  =  + GO 

for  all  values  of  *  and  ?/  which  are  connected  by  the  relation  6). 


APPLICATIONS  191 


It  is  clear  that,  if  y  —  /3  is  a  root  of  the  equation 

7)  *(y)  =  o, 

then  x  =  k/3  will  be  a  root  of  5). 

The  factor  k  is  arbitrary,  and  we  can  always  determine  it  so 
that,  on  dividing  equation  7)  through  by  Jc  : 

the  coefficient  of  y-  will  be  numerically  equal  to  unity  (provided 
that  p  =£  0) : 

i)  £=1         or         7c  =  Vp,  if    p>0; 

K'2 

ii)  P=-\     or         &  =  V^7,         if    p<°- 

In  this  way,  equation  5)  can  be  reduced  to  one  of  the  two 
forms 

a)  y4  +  ?/2  +  Ay  +  B  =  0; 

/8)  2/4  -  ?/-  +  -4»  +  5  =  0. 

If,  in  particular,  p  =  0  and  q  ^  0,  5)  can  be  reduced  to  the 
form 

y)  y*  +  y  +  B  =  Q. 

The  method  can  be  applied  to  any  algebraic  equation  whose 
|  second  term  is  lacking : 

xn  +  c2xn-2  +  c3a;n-3  +  ...  +  cn  =  0. 

EXERCISES 

1.    Replace  the  equation 

7a4  -  175a2  +  16x  +  10  =  0 

by  an  equation  of  the  type  /?),  and  state  precisely  the  relation 
>of  the  roots  of  the  second  equation  to  those  of  the  first. 


192  CALCULUS 

2.  Show  that,  if  in  the  equation 

aQxn  +  ayxn~l  +  •••  4-  an  =  0, 
where  a0  4=-  0  and  a„  =£  0,  the  transformation 

1 

y  =  - 

X 

is  made,  the  roots  of  the  new  equation, 

a„yn  +  «„-i2/R~l  +  •-  +  a0  =  0 
are  the  reciprocals  of  the  roots  of  the  given  equation. 

3.  If  on  transforming  equation  1)  by  2),  where  h  is  deter- 
mined by  4),  the  constant  term  in  the  resulting  equation  3), 
<f>(y)  =  0,  does  not  vanish,  the  further  transformation 

8)  y  =  -,         or         x  =  -  +  h, 

z  z 

will  carry  1)  into  an  equation  in  which  the  linear  term  is  lack- 
inS:  Az*  +  Bz*  +  D  =  0.       A  4=0,     Z>^0. 

The  theorem  holds  in  full  generality  for  an  algebraic  equa- 
tion of  any  higher  degree.     State  it  accurately. 

4.  Replace  the  equation 

tf  -  4a-3  -  6  x-  +  16  x  -  4  =  0 
by  an  equation  of  the  type 

Ay*  +  By3  +  Ctf-  +  D  =  0. 

Graphical  Treatment.     We  have  already  seen  that  the  cubic 
x3  +  px  +  q  =  0 
can  be  solved  graphically  by  cutting  the  standard  graph 

y  =  x3 

by  the  straight  line, 

y  =  -px-q. 

Since  the  general  cubic  can  be  reduced  by  the  transformation 
2)  to  a  cubic  of  this  type,  we  may  consider  the  general  problem 
of  the  graphical  solution  of  a  cubic  as  solved. 


APPLICATIONS  193 


To  obtain  a  similar  solution  for  the  general  biquadratic, 

9)  ax4  +  bx3  +  ex  +  dx  +  e  =  0,  a  =£  0, 

begin  by  reducing  it  to  one  of  the  three  forms  : 

i)  y4  +  y-  +  Ay  +  B  =  0; 

ii)  y4-y-  +  Ay  +  B  =  0; 

iii)  ?/4  +  A'/  +  B  =  0. 

An  equation  of  type  i) : 

x4  +  x+Ax  +  B  =  0, 

can  be  solved  graphically  by  cutting  the  standard  curve 

y  =  x4 
by  the  parabola 

y  =  —  a?2  —  At  —  B. 

A  similar  procedure  leads  to  a  solution  in  the  case  of  each 
of  the  other  two  types,  ii)  and  iii). 

TJie  Method  of  Curve  Plotting.  Let  the  coefficients  a,  e  in 
equation  9)  be  different  from  0.  By  means  of  Ex.  3,  p.  192,  the 
equation  can  be  reduced  to  one  of  the  following  type : 

Ax4  +  Bx3  +  Cx1  +  E  =  Q>. 

In  order  to  discuss  the  number  and  location  of  the  roots  of 
this  equation,  it  is  sufficient  to  plot  the  curve 

y  =  Ax4  +  Bx*  +  Cx°-  +  E. 

|  Since  all  the  maxima,  minima,  and  points  of  inflection  of  this 
curve  can  be  determined  by  means,  at  most,  of  quadratic  equa- 
tions, the  problem  is  readily  solved  in  any  given  numerical  case. 

EXERCISES 

Determine  the  number  of  real  roots  of  each  of  the  following 
equations,  and  locate  them  approximately. 

1.  3z4  +  8.t3  -90cc2+  100  =  0. 

2.  3  x4  +  8  a;3  -  90  x1-  +  500  =  0. 


194  CALCULUS 

3.  3a* +  833  — 90 »- +  1500  =  0. 

4.  Show  that  the  equation 

3a;4  +  4:e3  +  2a;2  +  l  =  0 
has  no  real  roots. 

How  many  real  roots  has  each  of  the  following  equations  ? 

5.  as5—  5x  — 1  =  0.  6.   a?  +  7z-l=0. 
7.   b3_  4a?  +  l  =  0.  8.    a,-3  -3a  — 2  =  0. 

9.   ^  _  z.  +  3  =  0  10    4^3  _  15  X2  +  12  x  +  1  =  0. 

11.    3a4  +  4x-3  +  6a2  -1  =  0.    12.    Sx4  -  4.T3  +  12x2  +  7  =  0. 

13.  How  many  positive  roots  has  the  equation 

6a4  +  8a-3  -  12a;2  -  24a;  - 1  =  0? 

14.  Has  the  equation 

3a*-8a*+12a?  +  l  =  0 

any  real  roots  ? 

15.  By  means  of  the  graph  of  the  function 

II  =  x3  +  px  +  q 
show  that  the  equation 

0?  +  px  +  q  =  0 
has 

(a)  1  real  root  when     -^  +  —  >  0 ; 

(b)  3  real  roots  when     —  +  --  <  0 ; 
v  )  27      4 

(c)  2  real  roots  when     i-  +  ^  =  0,     { p  and  7  not  both  0 } 

27      4 

(d)  1  real  root  when      l^  +  2!  =  0,     {^  =  9  =  0} 

27      4 

In  case  (c)  it  is  customary  to  count  one  of  the  roots  twice  j 
in  case  (d),  to  count  the  root  three  times. 


APPLICATIONS 


195 


16.  Extend  the  criterion  of  Ex.  15  to  the  ease  of  the  general 
cublc  ax3  +  bx2  +  ex  +  d  =  0. 

10.  Curve  Plotting.  We  will  close  this  chapter  by  consider- 
ing the  application  of  the  principles  set  forth  in  the  earlier 
paragraph  on  curve  plotting  (Chap.  Ill,  §  5)  to  some  interest- 
ing curves  of  a  more  complex  nature. 


1) 


Example  1.     To  plot  the  curve 

1  1 

y  = r  + 


1       X  +  1 


The  curve  is  obviously  not  symmetric  in  either  axis ;  but 
the  test  for  symmetry  in  the  origin  is  fulfilled,  since  on  replac- 
ing x  by  —  x  and  y  by  —  y  the  new  equation, 

1         .         1 


-y  = 


X  —  1         —  X  +  1 


is  equivalent  to  the  original   equation,  1).     Incidentally  we 
observe  that  the  curve  passes  through  the  origin. 

In  consequence  of  the  symmetry  just  noted  it  will  be 
sufficient  to  plot  the  curve  for  positive  values  of  x  and  then 
rotate  the  figure  about  the  origin  through  180°. 

To  each  positive 
value  of  x  but  one 
there  corresponds 
one  value  of  y. 
When  x  approaches 
1  as  its  limit  from 
iabove  (i.e.  always 
remaining  greater 
than  1),  y  becomes 
positively  infinite. 
Hence  the  line  x  =  1 
is  an  asymptote  for 
one  branch  of  the 
curve.  Fig.  61 


196  CALCULUS 

When  x  approaches  1  from  below,  y  becomes  negatively 
infinite,  and  hence  this  same  line,  x  =  l,  is  an  asymptote  for  a 
second  branch  of  the  curve. 

For  all  other  positive  values  of  x,  y  is  continuous. 

The  slope  of  the  curve  is  given  by  the  equation 

2)  d»  =  -(      1       + 


dx         V('--l)2     (»  +  l)2 

and  is  seen  to  be  negative  for  all  values  of  x  for  which  y  is 
continuous.  Thus,  in  particular,  the  curve  is  seen  to  have  uo 
maxima  or  minima,  or  in  fact  any  points  at  which  the  tangent 
is  horizontal. 

The  second  derivative  is  given  by  the  formula 

3)  *M  =  2(      X       +       1 

dx2        \(x-  l)3      0  +  1): 

When  x  >  1,  the  right-hand  side  of  this  equation  is  always 
positive,  and  so  the  curve  is  concave  upward  in  this  interval. 
Moreover,  it  is  evident  from  1)  that,  when  x  =  -\-cc,  y  ap- 
proaches 0  from  above,  and  so  the  positive  axis  of  x  is  also  an 
asymptote. 

In  the  interval  0  <  x  <  1,  the  second  derivative  is  surely 
sometimes  negative,  for  this  is  obviously  the  case  when  a 
is  only  slightly  less  than  1.  Is  d2y/dx2  always  negative  in 
this  interval  ?  If  not,  it  must  pass  through  the  value  0 
for  a  continuous  function  cannot  change  from  a  positive  tc 
a  negative  value  without  taking  on  the  intermediate  value 
0.*  Let  us  set,  then,  the  right-hand  side  of  equation  3' 
equal  to  0  and  solve: 

1    +^ 


(x  - 1)»    (x  +  ly 

*  How  must  the  graph  of  a  continuous  function  look,  which  is  some 
times  positive  and  sometimes  negative  ?  It  must  cross  the  axis  of  ab| 
scissas,  must  it  not  ?  At  the  point  or  points  where  it  crosses,  the  f  unctioi 
has  the  value  0. 


APPLICATIONS  197 

This  equation  is  equivalent  to  the  following : 
1  1 

(x-iy       o  +  i)3' 

Extracting  the  cube  root  of  each  side  of  this  equation,  we  have : 

1     = 1_ 

x -  1  x  +  1 

Clearing  of  fractions  we  find  : 

aj  +  l=-(»-l), 
or  2x  =  0. 

Hence  x  =  0  is  the  only  value  of  x  for  which  d2y/dx2  can 
vanish,  and  we  see  at  once  that  the  right-hand  side  of  3)  does 
vanish  for  x  =  0. 

We  have  thus  proven  that  the  continuous  function  3)  is  no- 
where 0  in  the  interval  0  <  x  <  1,  and  since  it  is  negative  in 
part  of  this  interval,  it  is  negative  throughout.  Hence  the 
curve  is  concave  downward  throughout  the  interval. 

It  is  now  easy  to  complete  the  graph.  The  curve  has  one 
point  of  inflection,  —  namely,  the  origin,  —  and  the  slope  there 
is,  by  2),  equal  to  —  2. 

EXERCISES 


Plot  the  following  curves  : 


1. 

V~3  +  X2' 

3. 

1    J    1 

y=x-2+x+2 

5. 

1^     1 

y=   +       i- 

X        05  +  1 

7. 

1 

X2 

9. 

1 

ar 

3x 
2.   y  = 


3  + a:2 

1  .       1 
4.    y  =  -  + 


x     x  —  1 

1 
6.  y  = 


8.  y  = 
10.   y  = 


-1 

1 


(i-xy 

l 

(x  +  iy' 


198  CALCULUS 

11.    y  =  x-\ 12.    y  =  x 

x  x 

A  ^ 

13.   y  =  — —  +  x2  —  2x.  14.   y  =  - 6x  —  x2. 

B      1-x  '       3+x 

11                                     1          1 
15.   y  = —  •  16.   y  =  ~- -> 

X  —  1       X  +  1  X       X—  1 

Example  2.     To  plot  the  curve 

4)  y2  =  x2  +  a,-3. 

We  observe  first  of  all  that  the  curve  is  symmetric  in  the 
axis  of  x.  It  is  sufficient,  therefore,  to  plot  the  curve  for  posi- 
tive values  of  y,  and  then  fold  this  part  of  the  curve  over  on 
the  axis  of  x.     The  curve  goes  through  the  origin. 

Unlike  the  examples  hitherto  considered,  this  curve  does 
not  permit  an  arbitrary  choice  of  x.  It  is  only  when  the  right- 
hand  side  is  positive  or  zero,  i.e.  when 

x~  +  x3  ^  0, 
or 

a^(l4-cc)^  0         or         x  >  -  1, 

that  there  will  be  a  corresponding  value  of  y  and  thus  a  point 
with  the  given  abscissa. 

Obviously,  the  curve  cuts  the  axis  of  x  at  the  origin  and  at 
the  point  x  =  —  1 .     We  have,  then,  essentially  two  problems  : 

i)  to  plot  the  curve  for  x  >  0 ; 
ii)  to  plot  the  curve  for  —  1  <  x  <  0. 

i)  When  x  >  0,  the  positive  value  of  y  is  given  by  the 
equation 


5)  y  =  xVl  +  x. 

Hence 

dx   2vr+» 


APPLICATIONS 


199 


For  positive  values  of  x  the  right-hand  side  of  this  equation 
is  always  positive,  and  hence  there  are  no  horizontal  tangents 
in  the  interval  under  consideration ;  the  slope  of  this  part  of 
the  curve  is  always  positive.  In  particular,  the  slope  at  the 
origin  is  unity : 

dy 

dx 


=  1. 


') 


The  second  derivative  has  the  value 

d*y       4  +  3 x_ 
dx-     4(1 +  a.)* 


The  right-hand  side  of  this  etpiation  is  always 
positive  in  this  interval,  and  thus  it  appears 
that  the  curve  is  concave  upward  for  all  posi- 
tive values  of  x. 


Fig.  62 


ii)  When  —  1  <  x  <  0,  the  positive  value  of  y  is  no  longer 
given  by  the  formula  5),  since  x  is  now  negative.*  In  the 
present  case, 


8) 

y  = 

—  a*Vl  +  x, 

and 
9) 

consequently 

dy  = 

dx 

2+ 3a 

2Vl  +  a;' 

10) 

d2y  _ 
dx2 

4  +  3.C 
4(1  +  a>)* 

The  first  derivative  will  vanish  if,  and 

only 

if, 

2 

-f-  3x  =  0, 

*  The  student  must  have  clearly  in  mind  the  definition  of  the  function 
expressed  by  the  y/  sign,  which  was  laid  down  in  Chap.  I,  §  1.  This  func- 
tion is  the  positive  square  root  of  the  radicand  ;  it  can  never  take  on  a 
negative  value. 


200 


CALCULUS 


It   is,  therefore,  important   to   determine   the   corresponding 
point  on  the  curve  and  draw  the  tangent  there : 

*l— I — (-i)Vw=^L.38. 

Two  other  important  points  for  the  present  curve  are  the 
origin  and  the  point  x  =  —l,  y  =  0.  At  these 
points  the  slope  has  the  following  values: 


dy 

dx 


—  _  1  •  "•'' 


Fig.  63         Draw  the  corresponding  tangents. 

From  the  expression  10)  for  the  second  derivative  it  is 
clear  that,  when  —  1  <  x  <  0,  the  right-hand  side  of  this 
equation  is  always  negative,  and 
hence  the  curve  is  concave  down- 
ward throughout  the  whole  in- 
terval in  question.  We  can  now 
draw  in  the  curve  in  this  interval, 
Fig.  63. 

The  curve  is  now  complete  above 
the  axis  of  x.  It  remains,  therefore, 
merely  to  fold  this  part  over  on  that 
axis.  The  entire  curve  is  shown  in 
Fig.  64. 

EXERCISES 


Fig.  (14 


Plot  the  following  curves  : 

l.   >/'  =  a*'2  —  x>.  2.  i/-  =  x  —  2x]  +  x5. 

3.  y-=(x—  a)-(Ax  +  B) 

Suggestion  :  Write  the  second  factor  in  the  form 

Ax  4-  B  —  A  (x  —  b).  where  b  =  — , 

A 


and  make  two  cases  :  i)  A  >  0 ;  ii)  A  <  0. 

case,  A  —  0. 


Discuss  the  omitted 


APPLICATIONS  201 

4.  y2  =  x-  —  xA.  5.  y2  =  x-  +  x*. 

Example  3.     To  plot  the  curve 
11)  y*  =  x(x-l)(x-2). 

The  curve  lies  wholly  in  the  regions 

0  <  x  <;  1         and         2  <;  x. 

It  is  symmetric  in  the  axis  of  x,  and  hence  it  is  sufficient  to 
plot  it  for  positive  values  of  ?/. 
The  function 

y  =  Vic(.x  —  l)(x  —  2) 

is  continuous  in  the  interval  0  <  x  <  1.  It  starts  with  the 
value  0  when  x  —  0,  increases,  and  finally  decreases  to  0  when 
x  =--  1. 

When  x,  starting  with  the  y 
value  2,  increases,  y,  starting 
with  the  value  0,  increases, 
always  remaining  positive,  and 
increasing  without  limit  as  x  be- 
comes infinite. 

So  much  from  considerations  of  continuity.  A  more  specific 
discussion  of  the  character  of  the  curve  can  be  given  by  means 
of  the  derivatives  of  the  function. 

The  slope  is  given  by  the  formula 


Fig.  65 


12) 

or 

13) 


2yC^='3x*-6x  +  2 
dx 


cly 


3^_6.r  +  2 


*»     2Vx(x-l)(x-2) 
The  slope  is  infinite  when  x  =  0  or  1 : 


dy\ 
dx\ 


dy 
dx 


=  CO. 


At  these  points,  the  tangent  is  vertical. 


202  CALCULUS 

The  slope  is  0  when 

3a.2_6a5+  2  =  0. 
The  roots  of  this  equation  are 

V3  V3 

The  first  of  these  values  does  not  correspond  to  any  point  on 
the  curve.  The  second,  x  =  .42,  yields  a  horizontal  tangent, 
the  ordinate  being 


i 
Fig.  66 


■"=\^=-62- 

Plot  this  point  and  draw  the  tangent.  From  the  above  dis- 
cussion on  the  basis  of  continuity  it  is  obvious  that  this  point 
must  be  a  maximum,  and  we  see  that  there 
are  no  other  maxima  or  minima.  But  it 
_x  is  not  clear  that  the  curve  has  no  points  of 
inflection  in  this  interval. 

To  treat  this  question,  compute  the  sec- 
ond derivative.  This  might  be  done  by  means  of  formula  13) ; 
but  it  is  simpler  to  use  12)  : 

2^  +  2^=6^-6, 

dx2        dx2 

il.r1  dx1 

Substitute  here  the  value  of  dy/dx  from  13)  and  reduce : 

14.  t r~j/  _  3  s4 -12  a8 +  12  a? -4 

)  V  dx-  4x(x-l)(x-2) 

And  now  we  seem  to  be  in  difficulty.  How  are  we  going  to 
tell  when  d2y/dx2  is  positive,  when  negative  ? 

First  of  all,  y  is  positive,  and  so  the  sign  of  d2y/dx2  will  be 
the  same  as  that  of  the  right-hand  side  of  the  equation. 

Secondly,  in  the  interval  in  question,  0  <  x  <  1,  the  denomi- 
nator is  positive. 


APPLICATIONS  203 

All  turns,  then,  on  whether  the  numerator,  i.e.  the  function 

15)  u  =  3^-12x3  +  12a;2  -4, 

is  positive  or  negative.  To  answer  this  question,  plot  the 
graph  of  the  function  15).  The  slope  of  the  graph  is  given  by 
the  equation 

16)  ^  =  12a?  -  36 x-  +  24,x=12x(x-  1)  (x  -  2). 
dx 

In  the  interval  in  question,  the  right-hand  side  of  this  last 
equation  is  always  positive.  Hence  u  increases  with  x  through- 
out the  interval  0  5^  x  <  1,  and  consequently  attains  its  great- 
est value  at  the  end-point,  x  =  1.     Here, 

W  |x=l  =  -  1. 

We  see,  therefore,  that  u  is  negative 
throughout  the  whole  interval  in  question, 
and  consequently  the  graph  of  1)  is  concave 
downward  in  this  interval. 

The  reasoning  by  which  we  determined  whether  u  is  pos- 
itive or  negative  is  an  excellent  illustration  of  the  practical 
application  of  the  methods  of  curve  plotting  which  we  have 
learned.  It  is  in  no  wise  a  question  of  the  precise  values  of 
u  which  correspond  to  x.  The  question  is  merely :  Is  u  posi- 
tive, or  is  it  negative?  Without  the  labor  of  a  single  com- 
putation involving  table  work  we  have  answered  this  ques- 
tion with  the  greatest  ease.  Such  questions  as  these  arise 
again  and  again  in  physics,  and  the  aid  which  the  calculus  is 
able  to  render  here  is  most  important. 

One  further  point.  It  may  seem  to  have  been  a  fluke  that 
we  were  able  to  factor  the  polynomial  in  16)  and  thus  simplify 
so  materially  the  further  discussion.  And  yet,  in  the  problems 
which  arise  in  practice,  —  the  problems  with  a,  pedigree,  —  just 
such  simplifications  as  this  present  themselves  with  great 
frequency. 


i 
Fig.  67 


204 


CALCULUS 


To  complete  the  graph,  it  remains  to  consider  the  interval 

2  ^  x  <  oo.     Since  , 

JL\     =oo, 
dx,z=2 

the  tangent  to  the  curve  is  vertical  at  the  point  where  the  curve 
meets  the  axis  of  x.  It  is  clear,  then,  that  the  curve  must  be 
concave  downward  for  a  while,  and  so  cPy/dx2  <  0  for  values 
of  x  slightly  greater  than  2.     This  is  verified  from  14),  since 

17)  «U=-4. 

On  the  other  hand,  when  x  is  large,  u  is  positive  and  (Vy/dx1 
is  positive.  Hence  the  curve  is  concave  upward.  There  must 
be,  therefore,  a  point  of  inflection  in  the  interval,  and  there 
may  be  several. 

From  14)  we  see  that  the  second  derivative  will  vanish  when 
and  only  when       ^  _  -^  +  12xi  _  4  =  0. 

The  problem  is,  then,  to  determine  the  number  of  roots  of 
this  equation  which  are  greater  than  2,  and  to  compute  them. 

Again,  it  is  a  question  of 
the  graph  of  15).  When 
x  >  2,  we  see   from  16) 

that      c^|     >  0 

dx\x>2 

Hence  u  steadily  increases 
with  x.  Now,  from  17), 
u  starts  with  a  negative 
value,  and  u  is  positive 
and  large  when  x  is  large. 
Hence  u  vanishes  for  just 
one  value  of  x  which  is 
greater  than  2.  Since 
u  |I=3  =  23,  this  root  is  seen 
to  lie  between  2  and  3. 
It  can  be  determined  to 
any  required  degree  of  accuracy  by  the  foregoing  methods  of 


Fin.  68 


APPLICATIONS  205 

this  chapter,  which  find  herewith  a  practical  application.     To 
two  places  of  decimals  it  is  2.47. 

EXERCISES 

Plot  the  following  curves  : 

1.    y  =  x1  —  x.  2.  y  =  x  —  x3. 

3.    if  =  x3  +  x.  4.  if  =  1  —  a;4. 

5.    if-  =  (a;2  -  1)(«'  -  4).  6.  if  =  (1  -  x")(x  -  4). 

-      -       '    1  -  1/2  1 


.r  —  a; 

9. 

2                X 

1  —  a; 

11. 

,          a;2 

1  +  a*" 

13. 

9             #' 

^          8,-1 

15. 

?/'-  =  a;3  —  4  a;'  +  3  x. 

17. 

y  =  sin  x—  sin  2  a;. 

19. 

7/  =  cos  x  —  cos  2x. 

it 

(a;2—  l)(z2  —  4) 

10 

f 

£C 

1+JB 

12 

y' 

a52 

1  -  a;2 

14 

f- 

cc1 

1  +  X 

16. 

y 

=  sin#  +  sin  2  a;. 

18. 

V 

=  cos  x  +  cos  2  a;. 

20. 

y 

=  x  -+-  sin  x,  0  <  x  ^  7r 

CHAPTER    VIII 
THE   INVERSE   TRIGONOMETRIC   FUNCTIONS 

1.   Inverse  Functions.     Let 

(1)  »«/(«) 

be  a  given  function  of  x,  and  let  us  solve  this  equation  for  x  as 
a  function  of  y : 

(2)  a  =  *(y). 

Then  4>{y)  is  called  the  inverse  function,  or  the  inverse  of  the 
function  f(x).     Thus  if  f(x)  =  x3,  we  have 

y  =  x*. 

Hence  x  =  s/y, 

and  <£(//)  is  here  the  function  -yjy. 

When  the  given  function  is  tabulated,  the  table  also  serves 
as  a  tabulation  of  the  inverse  function.  It  is  necessary  merely 
to  enter  it  from  the  opposite  direction.  Thus,  if  we  have  a 
table  of  cubes,  we  can  use  it  to  find  cube  roots  by  simply  re- 
versing the  roles  of  the  two  columns. 

In  the  same  way,  the  graph  of  the  function  (1)  serves  as  the 
graph  of  the  function  (2),  provided  in  the  latter  case  we  take 
y  as  the  independent  variable,  and  x  as  the  dependent  variable, 
or  function. 

The  graph  of  the  inverse  function,  plotted  with  x  as  the  in-  K 
dependent  variable,  can  be  obtained  from  the  former  graph  as  $ 
follows.  Make  the  transformation  of  the  plane  which  is  de-  -ft 
fined  by  the  equations  : 

(3)  X',  =  y>\  or  "£ 
y'  =  x,  j                            y  =  x'. 

206 


THE   INVERSE   TRIGONOMETRIC   FUNCTIONS      207 

It  is  easy  to  interpret  this  transformation.  Any  point,  whose 
coordinates  are  (x,  y),  is  carried  over  into  a  point  (pa',  y')  situ- 
ated as  follows  :  Draw  a  line  L  through  the  origin  bisecting  the 
angle  between  the  positive 
axes  of  coordinates.  Drop 
a  perpendicular  from  (x,  y) 
on  L  and  produce  it  to  an 
equal  distance  on  the  other 
side  of  L.  The  point  thus 
determined  is  the  point 
(%',  y').  The  proof  of  this 
statement  is  immediately 
evident  from  the  figure. 

Thus  it  appears  that  the 
transformation  (3)  can  be 
generated  by  rotating  the 
plane  about  L  through 
180°. 

The  transformation  is  also  spoken  of  as  a  reflection  in  L, 
since  if  a  plane  mirror  were  set  at  right  angles  to  the  plane  of 
(x,  y)  and  so  that  the  line  L  would  lie  in  the  surface  of  the 
mirror,  the  image  of  any  figure,  as  seen  in  the  mirror,  would 
be  the  transformed  figure. 

Monotonia  Functions.  A  function,  f(x),  is  said  to  be  mono- 
tonic  if  it  is  single-valued  and  if,  as  x  increases,  f(x)  always 
increases,  or  else  always  decreases.  We  shall  be  concerned  only 
<with  functions  which  are,  in  general,  continuous.  It  is  obvious 
that  the  inverse  of  a  monotonic  function  is  also  monotonic. 

A  given  function, 

:can  in  general  be  considered  as  made  up  of  a  number  of  pieces, 

each  of  which  is  monotonic  in  a  certain  interval.*     Thus  the 

function 

|(4)  y  =  & 


Fig.  69 


*  There  are  functions  which  do  not  have  this  property  ;   but  they  do 
aot  play  an  important  role  in  the  elements  of  the  Calculus. 


208  CALCULUS 

can  be  taken  as  made  up  of  two  pieces,  corresponding  respec- 
tively to  those  portions  of  the  graph  which  lie  in  the  first  and 
the  second  quadrants,  the  corresponding  intervals  for  x  being 

here  -  oo  <  .^  0,  0  <  x  <  oo. 

Each  of  the  pieces,  of  which  fix)  is  made  up,  has  a  monotonic 
inverse,  and  thus  the  function  <j>(x)  inverse  to  f(x)  is  repre- 
sented by  a  number  of  monotonic  functions. 

In  the  example  just  cited,  the  inverse  function  is  multiple- 
valued  : 

(5)  y  =  ±  Va;. 

But  one  of  the  two  pieces  into  which  the  original  function  was 
divided  yields  the  single-valued  function 

(6)  y  =  Va, 

the  so-called  principal  value  of  the  multiple-valued  function 

(5) ;  the  other,  /— 

y  =  -Vx, 

the  remainder  of  (5). 

The  derivative  of  a  monotonic  function  cannot  change  sign ; 
but  it  can  vanish  or  become  infinite  at  special  points.     Thus 

y  =  Va2  —  x'2,  0  ^  x  <^  a, 

is  a  decreasing  monotonic  function.  Its  derivative  is,  in  gen- 
eral, negative  ;  but  when  x  =  0,  it  vanishes,  and  when  x  =  a, 
it  becomes  infinite. 

Differentiation  of  an  Inverse  Function.  The  function  <f>(x) 
inverse  to  a  given  function  f(x)  can  be  differentiated  as  follows! 
By  definition^ the  two  equations 


e 


(7)  y  =  <f>(x)  and  x=f(y) 

are  equivalent ;  they  are  two  forms  of  one  and  the  same  rela- 
tion between  the  variables  x  and  y.     Their  graphs  are  identical* 
Take  the  differential  of  each  side  of  the  second  equation : 

dx  =  dJ\y)  =  DyJ\y)-dy. 


THE   INVERSE   TRIGONOMETRIC   FUNCTIONS      209 


Hence 

(8) 


dy  _       1 

dx~  DJ(y) 


To  complete  the  formula,  express  the  right-hand  side  of  (8) 
in  terms  of  x  by  means  of  (7). 

2.  The  Inverse  Trigonometric  Functions.  The  inverse  trigo- 
nometric functions  are  chiefly  important  because  of  their 
application  in  the  Integral  Calculus.  They  are  defined  as 
follows. 

(a)   TJie  Function  sin-1#.     The  inverse  of  the  function 

(1)  y  =  sin  x 

is  obtained  as  explained  in  §  1  by  solving  this  equation  for  x 
as  a  function  of  y,  and  is  written  : 

(1')  x  =  sin-1  y, 

read  "  the  anti-sine  of  ?/."  *  In  order  to 
obtain  the  graph  of  the  function 

(2)  ?/  =  sin-1  a; 

we  have,  then,  merely  to  reflect  the  graph 
of  (1)  in  the  bisector  of  the  angle  made 
by  the  positive  coordinate  axes.  We 
are  thus  led  to  a  multiple-valued  func- 
tion, since  the  line  x  =  x'(—  1  <:  x'  <  1) 
cuts  the  graph  in  more  than  one  point, 
—  in  fact,  in  an  infinite  number  of  points. 
For  most  purposes  of  the  Calculus,  how- 
ever, it  is  allowable  and  advisable  to  pick  Fig.  to 

*  The  usual  notation  on  the  Continent  for  sin-1  x,  tan-1  x,  etc.,  is  arc  sin  x, 
I  arc  tan  x,  etc.  It  is  clumsy,  and  is  followed  for  a  purely  academic  reason  ; 
namely,  that  sin_1x  might  be  misunderstood  as  meaning  the  minus  first 
|  power  of  sin  x.  It  is  seldom  that  one  has  occasion  to  write  the  recipro- 
cal of  sin  x  in  terms  of  a  negative  exponent.  When  one  wishes  to  do  so, 
call  ambiguity  can  be  avoided  by  writing  (sinx)-1. 


210  CALCULUS 

out  just  one  value  of  the  function  (2),  most  simply  the  value 
that  lies  between  y  =  —  tt/2  and  y  =  -+-  tt/2,  and  to  understand 
by  sin-1  a;  the  single-valued  function  thus  obtained.  This  de- 
termination is  called  the  principal  value  of  the  multiple-valued 
function  sin_1x.  Its  graph  is  the  portion  of  the  curve  in 
Fig.  70  that  is  marked  by  a  heavy  line.  This  shall  be  our 
convention,  then,  in  the  future  unless  the  contrary  is  explic- 
itly stated,  and  thus 

(3)  y  =  sin  "  x 

is  equivalent  to  the  relations  : 

(3')  x  =  sin//.  _*<y<|. 

In  particular, 

sin-1 0  =  0,         sin-1l=^3         sin-*(- 1)=-|. 

The  student  should  now  prepare  a  second  plate,  showing 
the  graphs  of  the  three  functions  sin-1  x,  cos-1  x,  tan-1  x.  Place 
the  first  in  the  upper  left-hand  corner  of  the  sheet ;  the  second, 
in  the  upper  right-hand  corner ;  and  the  third  on  the  lower 
half-sheet.  All  of  these  curves  can  be  ruled  from  the  templets. 
Use  a  hue  lead-pencil ;  then  mark  in  the  principal  value  of 
the  function  in  a  clean,  firm  red  line.  Also  mark,  in  each  fig- 
ure, all  the  principal  points,  as  is  done  in  Fig.  70  of  the  text. 

Differentiation  o/sin-1a;.     Ln  order  to  differentiate  the  func-    I; 

tion  .     . 

y  =  sin  l  x, 

make  the  equivalent  equation, 

x  =  sin  y 

the  point  of  departure.     Then 

dx  =  d  sin  y  =  cos  y  dy. 

Hence  -  = 

dx     cos  y 


THE   INVERSE  TRIGONOMETRIC  FUNCTIONS     211 


The  right-hand  side  of  this  equation  can  be  expressed  in 
terms  of  x  as  follows.     Since 

sin2  y  +  cos2  y  =  1 

and  since  sin  y  =  x,  we  have  , 

cos2  y  =  1  —  x2,  cos  y  =  ±  Vl  —  x2. 

We  have  agreed,  however,  to  understand  by  sin-1  a;  the 
principal  value  of  this  function.  Hence  y  is  subject  to  the 
restriction  :  —  tt/2  ^y^  tt/2,  and  consequently  cos  y  is  posi- 
tive (or  zero).  We  must,  therefore,  take  the  upper  sign  before 
the  radical, #  the  final  result  thus  being : 


(4) 

or 


d    .   _.  1 

— sin  1x  =  — 
dx  Vl- 


d  sin  1  x  = 


dx 


vr 


(6)   The  Function  cos  lx.     The  treatment  here  is  precisely 
similar.     The  definition  is  as  follows  : 


(5) 


y  =  COS   1  X 


il- 


cc  =  cos  y, 


(read  :  "  anti-cosine  x  "). 

The  graph  of  the  function  cos_1x  is  as 
shown  in  Fig.  71.  Like  sin_1x,  this  function 
is  also  infinitely  multiple-valued.  A  single- 
valued  branch  is  selected  by  imposing  the 
further  condition 

This  determination  is  known  as  the  principal 
t value  of  cos-1  a: : 


(6) 


y  =  cos-1  x, 


O^y^ir. 


*  Geometrically  the  slope  of  the  portion  of  the  graph  in  question  is 
talways  positive,  and  so  we  must  use  the  positive  square  root  of  1  —  x2. 


212  CALCULUS 

It  will  be  understood  henceforth  that  the  principal  value  is 
meant  unless  the  contrary  is  explicitly  stated. 

In  preparing  the  graph  of  this  function,  mark  the  principal 
value  as  a  tirm  red  line. 

To  differentiate  the  function  cos_1#,  use  the  implicit  form 
of  equation  (5) : 

x  =  cos  y. 
Hence 

dx  =  (( cos  y  =  —  sin  y  dy 

and  1 


dx         sin# 

For  the  principal  value,  sin  y  is  positive,  and  hence 

(()  —cos  1x  = ' 

dx  Vl  -  x- 

or 

(, ')  t/cos  lx  =  — 


Vl  -  a2 

The  principal  values  of  the  functions  sin-1  x  and  cos-1  a;  are 
connected  by  the  identical  relation: 

(8)  sin-1  x  -f  cos-1  x  =  ^  • 

Li 

By  means    of   this    relation,    the   differentiation   of   cos-1# 
could  have  been  performed  immediately. 

(c)   The  Function  tan-1  a-.     Here,  the  definition  is  as  follows  : 

(9)  y  =  tan-1  a;  if  a:  =  tan?/, 

(read  :  "  anti-tangent  x  "). 

The  principal  value  is  defined  as  that  determination  which 
lies  between  —  tt/2  and  7r/2  : 

(10)  y  =  tan-ia,  -|  <y<"- 

In  preparing  the  graph  of  this  function,  mark  the  principal 
value  as  a  firm  red  line. 


THE   INVERSE   TRIGONOMETRIC   FUNCTIONS      213 

y, 


3  7T 

o 

Fig.  72 

To  differentiate  tan-1  a;  use  the  implicit  form  (9).     Hence 

dx  =  d  tan  y  =  sec2  y  dyy 

dx     sec2  y 
Since  sec2  ?/  =  1  +  tan2  y 

and  tan  y  =  x,  it  follows  that 

(11) 
or 

(12)  dtan-1* 


—  tan-1  x  = , 

dx  1+x* 


dx 


l  +  x1 
(d)  TJie  Function  cot-1  x.     Here,  the  definition  is : 

(13)  y  =  cot-1  x  if  x  —  cot  y, 

(read  :  "  anti-cotangent  x  "). 

The  principal  value  is  chosen  as  that  one  which  lies  between 
0  and  7r : 

(14)  y  =  cot-1  x,  0<y<Tr. 


214  CALCULUS 

The  differentiation  can  be  performed  as  in  the  case  of  the 
function  tan-1  x,  but  still  more  simply  by  means  of  the  identi- 
cal relation  connecting  the  principal  values  of  tan-1  a;  and 
cot-1  x : 

(15)  tan-1  x  +  cot-1  x  =  - . 
Hence 

(16)  Acolr1*^ L_ 

v    J  dx  1  +  z2' 

or 

(17)  dcot-1a;  =  --^-. 

1  +x2 

It  is  well  for  the  student  to  make  a  graph  of  this  function, 
also,  drawing  in  the  principal  value,  as  usual,  in  red. 

The  following  identity  holds  for  positive  values  of  x,  when 
the  principal  values  of  the  functions  are  used  : 

(18)  tan"1-  =  cot"1  x,  0<x. 

x 

For  negative  values  of  x  it  reads  : 

(18')  tan"1-  =  cot"1  x  -  w.  x<0. 

x 

Remarks.  The  other  inverse  trigonometric  functions,  sec_1#, 
esc-1  a;,  can  be  treated  in  a  similar  manner.  They  are,  how- 
ever, without  importance  in  practice.  Their  principal  values 
cannot  be  defined  by  means  of  a  single  continuous  curve. 
The  graph  necessarily  consists  of  more  than  one  piece  ;  it  is 
most  natural  to  take  it  as  consisting  of  two  pieces. 

Corresponding  to  the  Addition  Theorem  for  each  of  the 
trigonometric  functions,  there  are  functional  relations  for  the 
inverse  trigonometric  functions.     Thus,  for  tan-1  x : 

(19)  tan"1  u  +  tan-1  v  =  tan"1  *  +  v  ■ 

1  —  uv 

These    relations,    however,    are    not   always    true    when   the 
principal  value  of  each  of  the  functions  is  taken,  and  for  this 


THE   INVERSE   TRIGONOMETRIC   FUNCTIONS     215 

reason  it  is  usually  better  not  to  employ  them.  If,  however, 
in  a  particular  case,  u  and  v  are  each  numerically  less  than 
unity,  the  principal  values  can  be  used  throughout  in  (19). 

3.  Shop  Work.  The  student  will  now  add  to  his  list  of 
Special  Formulas  the  four  new  formulas  of  this  chapter.  The 
list  of  formulas  of  differentiation  is  now  complete.  It  reads 
as  follows. 

Special  Formulas  of  Differentiation 

1.  dc=0. 

2.  d  xn  =  nxn~l  dx. 

3.  d  sin  x  =  cos  x  dx. 

4.  d  cos  x  =  —  sin  x  dx. 

5.  cZtan  x  =  sec'  x  dx. 

6.  d  cot  x  =  —  esc'  x  dx. 


8. 
9. 

10. 
11. 
12. 
13. 


dlogx 

X 

dex 

=  exdx. 

dax 

=  aT  log  a  dx. 

d  sin-1  x 

dx 

Vl-a2 

d  cos-1 x 

dx 

VI  -  a'2 

d  tan-1  x 

dx 

1  +  X1 

d  cot-1  x  ■■ 

dx 

1  +  X2 


It  is  important  that  the  student  gain  facility  in  the  use  of 
the  new  results. 


216  CALCULUS 

Example  1.     Differentiate  the  function 


Let 
Then 


Hence        — 


II 

= 

cos-1  -  , 
a 

X 

y=— 

a 

u 

=  cos-1  y, 

du 

=  d  cos-1 y 

dy 

VI- 

V 

dy 

ll.r 
a 

dx 

a 

a  >  0. 


dy  a  dx 


VI  -f 


and,  finally, 


cZcos    -  = 


a  Va2  -  x* 

In  abbreviated  form, 

4-) 

,        .x  \aj  dx 

d cos-1  -  = 


a  L      /xV  Va2 


f~® 


Example  2.     Differentiate  the  function 


f       -,223  +  1 

a  =  tan  ' 

3 


Here, 

r2z  +  l 


d 

o  \dx  odx 

du  = 


±      /2a;  +  lY~10  +  4x'  +  4a-2      5  +  2x  +  2ce2) 
[     3     J  9 


du  _  3 

dx~  5  +  2x  +  2x2 


THE   INVERSE   TRIGONOMETRIC   FUNCTIONS     217 

The  student  should  notice  that  the  method  used  in  the  text 
for  deriving  the  fundamental  formulas  of  differentiation  is  not 
to  be  repeated  in  the  applications.  It  is  these  formulas  them- 
selves that  should  be  used.     Thus,  to  solve  Ex.  1  by  writing 

cos  u  =  - 
a 

and  then  differentiating  would  be  logically  irreproachable,  but 
bad  technique. 

EXERCISES 

Differentiate  each  of  the  following  functions. 

•   _,  ■>'  du  1  . , .     .    n 

1.  u  =  sin  !-• 

a 

2.  M  =  tan-1-- 

a 

3.  u  =  cot-1-- 

a 

4.  u  =  sin_1(n  sin  x). 

5.  u  =  cos  l  ■ • 


dx 

Va2 

)1L 

' 

du 

1 

dx 

a2  + 

X1 

du  = 

ol 

dx 

-f  X1 

du 

l  cos  X 

Vi 

dx 

—  n2  sin2 

X 

du  _ 

1 

dx     V3+2a-a;2 


•  -,2a; -1 

6.    m  =  sin  ' ■ 

V2 


il. 


m  =  cot 


10.    ?« =  tan~ 


1-x' 
11.    m  =  tan_1f  a;  - 


1  —  3  a;2 


a:  —  a 


12.    rt  =  sin-1  - 

x 


7 

u  =  cot  l ■ 

b 

9 

u  =  tan-1-- 

X 

du         2 

dx      1  +  a;2 

du          3 

dx     1  +  or2 

13. 

w  =  cos  i ■  • 

x  -\-a 

14.    u  =  sin-1  (2 x^T^x^j.  —  =        2    -,    \\x<—  ■ 

dx      Vl-a;2  V2 


8 

CALCULUS 

15. 

t  =  COS  l  -  • 

2 

—  =  -2  sin  2* 
dt 

16. 

3 

ds 

—  =3  cos  3  t. 
dt 

17.     t  =  cos-1  -+-/.  —  =  — n  sin  n(t  —  y)- 

n      7  eft  V        7; 

tan-1  :c 


18.    «  =  x  sin-1  x.  19.    a  = 

a; 

1  (/"  ' 

20.  «  = —  =  — ==== • 

Sin"1  x  dx      Vi  _  xi  (siu-i  xy2 

r    —  (X  X  —  ft 

21.  ?t  =  a  cos~i  —    — •  22.    M  =  tan_1 

a  .c  -f-  a 

23.     it  =  COl    '  24     W  =  Sill   r ! — • 

bx  —  a  bx  +  a 


/  .,                           , "                </»       V«2  —  (X2  A 

25.    tt=Va;       a       a  cos  '  ■  = ,  a  >  0. 


i  ■'■  ,  V  a2  —  x'1  da      Va'2  —  a;2       ^   ^ 

26.  ?^  =  siu1--( — = ,  a  >  0. 

a  a;  da;  .<■' 

27.  M=tan-^2tan->k 


28.    it  =  tan-i(3  tan  0) 


dx     5  — 3  cos  re 

'In  3 


(/#      5  —  4  cos  2 1 


4.  Continuation.  Numerical  Computation.  By  moans  of 
the  Tallies  the  numerical  value  of  any  of  the  functions  of  this 
chapter  can  l>e  determined  when  a  specific  numerical  value  hai 
been  chosen  for  the  independent  variable.  It  is,  however,  an 
important  aid  to  ease  and  security  in  such  computations  to  be 
able,  in  advance,  to  make  sure  of  the  early  significant  figures 
and  the  location  of  the  decimal  point.  There  are  two  impor- 
tant geometrical  methods  for  achieving  this  end.  One  is  the 
representation  of  the  trigonometric  functions  by  suitable  lines 
connected  with  the  unit  circle;  the  other  consists  in  the  graphs 
introduced  above,  in  §  2. 


THE   INVERSE  TRIGONOMETRIC  FUNCTIONS     219 

First  of  all,  however,  it  should  be  pointed  out  that  there  are 
two  distinct  problems.  One  is  to  find  all  values  of  x  which 
satisfy  such  equations  as 

(a)  sin  x  =  .2318  ; 

(6)  cos  x  =  -  .4322  ; 

(c)  tan  x  =  -  1.4861. 

The  other  is  to  find  the  principal  value  of  an  inverse  trigono- 
metric function ;  for  example, 

sin-i.2318  ;  cos"1  ( -  .4322) ;  tan~i  ( -  1.4861) 

The  methods  of  treating  these  problems  are  identical. 

First  Geometric  Method.  Equations  (a),  (b),  (c)  can  be 
solved  graphically  by  the  aid  of  the  unit  circle  representation 
with  an  error  corresponding  to  a  degree  or  two,  the  results 
being  expressed  in  radians  if  the  problem  comes  from  the 
Calculus. 

For  example,  consider  equation  (b).  The  student  should 
provide  himself  with  an  accurately  drawn  circle  of  his  own 
construction,  executed  on  the  accurate  centimeter-millimeter 
paper  commercially  procurable  ;  the  radius  of  the  circle  being 
10  cm.  and  its  center  at  a  principal  intersection  of  the  rulings. 

To  solve  equation  (6),  he  will  lay  a  straight-edge  on  his 
plate,  parallel  to  the  secondary  (or  ?/-)  axis  and  at  a  distance 
of  4  cm.,  3^  mm.  to  the  left  of  that  axis.  Marking  the  two 
points  of  intersection  of  the  straight-edge  with  the  circle  by 
fine  pencil  lines  easily  erased,  he  now  measures  one  of  the 
acute  angles  involved  by  means  of  his  protractor  and  thus 
determines  the  two  solutions  of  (6)  lying  between  0°  and  360° 
correct  to  minutes  or  thereabouts.  By  aid  of  the  Tables  the 
values  can  at  once  be  converted  into  radian  measure. 

Arithmetic  Solutions.  From  the  figure  before  him  the  stu- 
dent now  sees  clearly  a  right  triangle,  one  leg  of  which  is 
known.  The  determination  of  the  angle  he  needs  is  merely  a 
problem  in  the  solution  of  a  right  triangle  by  the  tables,  and 


220  CALCULUS 

he  proceeds  to  carry  this  work  through  to  the  degree  of  accu- 
racy which  the  tables  permit. 

Equations  (a)  and  (c)  are  treated  in  a  similar  manner.  The 
point  of  this  method  is  that  the  student  is  trained  to  visualize 
a  figure,  and  not  to  try  to  remember  a  table  that  looks  like 

sin  J  -f  +  —  — . 

For,  such  tables  vanish  in  a  short  time,  and  when  the  student 
needs  his  trigonometry  in  later  work,  he  is  helpless. 

In  terms  of  the  inverse  (unctions,  this  first  problem  consists 
in  finding  all  the  values  of  the  multiple-valued  function  cos-1  a; 
for  the  value  of  the  variable,  x  =  —  .4322. 

Second  Geometric  Method.  This  method  consists  in  reading 
off  from  the  graph  the  two  values  which  ,  een  0  and  L*  tt, 

and  then  adding  to  these  arbitrary  positive  or  negative  multi- 
ples of  2  ir. 

The  graph  suggests,  moreover,  how  to  determine  these  values 

arithmetically  by  the  aid  of  a  table  of  sines  or  cosines  of  angles 

of  the  first  quadrant.     It  also  suggests  a  further  refinement  of 

the  graphical  method,  of  which  the  student  will  do  well  to 

avail   himself,  —  namely,   this.     Let  him   make   an    accurate 

graph  of  the  function 

y  =  sin  x 

on  cm.-mm.-paper,  taking  10  cm.  as  the  unit  and  measuring 
the  angle  in  radians,  x  ranging  from  0  to  ir/2.  This  half-arch 
supplements  the  four  graphs  of  the  functions  sin  x,  cos  x,  sin-1  aj, 
cos-1  x  and  serves  as  a  3-place  table  for  determining  their  values 
(with  a  possible  error  of  two  or  three  units  in  the  third  place). 

To  sum  up,  then,  there  are  two  geometric  methods ;  1)  the 
unit-circle  method  ;  2)  the  graphs  of  the  functions,  the  latter 
being  supplemented  by  the  10-cm.  graph  just  described.  Either 
of  the  geometric  methods  suggests  how  to  use  the  tables  correctly 
and  affords  an  altogether  satisfactory  check  on  the  tables. 

When  the  accurately  drawn  graphs  are  not  at  hand,  free- 
hand drawings  indicate  clearly  how  to  use  the  tables  with 
security  and  accuracy. 


THE   INVERSE  TRIGONOMETRIC  FUNCTIONS     221 

EXERCISES 

1.  Determine  both  in  degrees  and  radians  all  values  of  x 
which  satisfy  the  above  equations  (a),  (&),  (c),  using  each  time 
all  of  the  geometric  methods  set  forth,  and  also  the  tables. 

2.  Find  the  value  of  each  of  the  following  functions.  It  is 
understood  that  the  principal  value  is  meant.  Use  first  the 
method  of  the  graphs.  Then  determine  from  the  tables. 
Check  by  unit-circle  and  protractor. 

i)  sin-i(-  .1643);  ii)  cos"1  (.6417) ; 

iii)  tan~i(-  2.8162). 

3.  By  means  of  a  free-hand  drawing  of  the  graph  estimate 
the  value  of  each  of  the  following  functions.  Remember  that 
a  curve  recedes  from  its  tangent  very  slowly  near  a  point  of 
inflection. 


a)  shr1 .113 ; 

b)  tan"1  (-.214); 

c)  cos"1. 172; 

d)  tan-1  (-7.4); 

e)  cot-1  (-.152); 

/)  cos-1  (-.998); 

g)  sin"1  (-.21); 

h)  sin-i.89; 

i)  tan-1 5.2; 

j)  cot-i7.3; 

k)  cos"1  (-.138); 

0  sin"1  (-.138). 

In  what  cases  is  your  error  large ;  in  what,  small  ? 

5.    Applications.     The  inverse  trigonometric  functions  afford 
a  convenient  means  of  solving  the  following  problem  in  Optics. 

A  ray  of  light  is  refracted  in  a 
prism.  Show  that  its  deviation  from 
its  original  direction  is  least  when 
the  incident  ray  and  the  refracted 
ray  make  equal  angles  with  the 
faces  of  the  prism. 

The  study  of  this  problem  has  a  FlG  73 

vivid  interest  for  the  student  who 

has  seen   the   laboratory  experiment   of   admitting  a  ray  of 
sunlight  into  a  darkened  room,  allowing  it  to  pass  through 


222  CALCULUS 

a  prism,  thus  being  refracted,  and  throwing  it  finally,  dis- 
persed, on  a  screen. 

Let  AP  be  the  incident  ray ;  PQ,  its  path  through  the 
prism ;  and  QB  the  ray  which  emerges.  Then  the  deflection 
of  PQ  is  obviously  6  —  <f>  and  the  further  deflection  of  QB  is 
6'  —  <f>' ;  so  that  the  total  deflection,  u,  is : 

(1)  u  =  e-4,  +  0'-<t>' =  6  +  6' -(<!>  + #). 

On  the  other  hand,  the  sum  of  the  angles  of  the  triangle 
PDQ  is  /  \      /  \ 

„.(!_♦)+(=_♦')+.. 

Hence 

(2)  $  +  <*>'=  a. 

We  can,  therefore,  write  (1)  in  the  form : 

(3)  u  =  6+6'  -a. 

This  is  the  quantity  it  is  desired  to  make  a  minimum.  6  and 
6'  are,  however,  connected  by  a  relation  which  can  be  obtained  as 
follows.     We  have  by  the  law  of  refraction  (cf.  Chap.  V,  §  7) : 

/)N  sin0  sin6' 

(4) =  n, =  n. 

sin  cf>  sm  <j>' 

Let  v  =  1/n.     Then 

(5)  sin<£  =  vsin#  or  <f>  =  sin-1  (v  sin  0). 
Similarly, 

(6)  sin  <f>'  =  v  sin  6'         or         <£'  =  sin-1  (v  sin  6'). 

Substituting  these  values  of  <£  and  <f>'  in  equation  (2)  we  have 
the  desired  relation : 

(7)  sin-1  (v  sin  6)  +  sin-1  (v  sin  6')  =  «. 

Our  problem  now  is  completely  formulated  ;  it  is  :  To  make 
the  function  u  given  by  (3)  a  minimum,  when  6  and  6'  are  con- 
nected by  (7) : 

11  =  6  +  6'-  a, 

(8)  \ 

sin-1  (v  sin  6)  +  sin-1  (v  sin  6')  =  a. 


THE  INVERSE  TRIGONOMETRIC  FUNCTIONS     223 

Take  0  as  the  independent  variable.     Then 
du_1      dP 

and  the  condition 

clu      A  „  dd'  ., 

—  =  0  gives  — -  =  —  1. 

dd  8  cW 

Next,  take  the  differential  of  each  side  of  the  second  equa- 
tion (8) : 

d  (v  sin  0)  d  (v  sin  6')      _  ^ 

Vl  -  v2  sin2  0      Vl  -  v2  sin2  0' 
or 

v  cos  0  dd  v  cos  ff  dd'     _  ^ 

Vl  -  v2  sin2  0      Vl  -  v2  sin2  0' 
Hence 
/-.  «v  cos  0  , cos  0'         /^^  —  0 

Vl  -  v-  sin-  0      Vl-v2sin20'U0 

But  d0'/d0  =.—  1.     Consequently 

,-...,.  cos  6  cos  6' 


Vl  —  v2  sin2  0      Vl  —  v!  sin2 


One  solution  of  this  equation  is  0  =  0',  —  the  solution  de- 
manded by  the  theorem.  But  conceivably  there  might  be 
other  solutions,  and  then  it  would  not  be  clear  which  one  of 
them  makes  u  a  minimum.  We  can  readily  show,  however, 
that  equation  (11)  has  no  further  solutions.     Square  each  side : 

cos2  0  cos2  0' 

1  —  v2  sin2  0      1  —  v-  sin2  0' 

i  Clear  of  fractions  and  express  each  cosine  in  terms  of  the  sine : 

(1  -  v2  sin2  0')(1  -  sin-  0)  =  (1  -  v-  sin2  0)(1  -  sin2  00- 

Multiply  out  and  suppress  equal  terms  on  the  two  sides  : 

—  sin2  0  —  v2  sin2  0'  =  —  sin2  d'  —  v  sin2  0, 

(v2  -  1)  sin2  0  =  (v2  -  1)  sin2  0'. 


224  CALCULUS 

Hence 

sin2  0  =  sin:  6',  sin  6  ==  sin  0', 

and  consequently  the  only  angles  of  the  first  quadrant  which 
can  satisfy  (11)  are  equal  angles,  6  =  6'. 

From  (5)  and  (6)  it  follows  that  <j>  =  <f>'.     Hence,  from  (2) 

d>  =  - ,         and  so 

u  =  2  sin-1  (  n  sin  - 
V         2 

That  u  is  a  minimum,  is  clearly  indicated  by  the  labora- 
tory experiment.  It  can  be  proven  analytically  as  follows. 
From  (9) 

d  "  d2d' 

(W-  ~     dB1  ' 

Differentiate  (10)  as  it  stands;  then,  after  the  differentia- 
tion, set  dd'/cW  =  —  1  and  6  =  6'.     It  is  seen  at  once  thaf 

M.  <  0,  hence  **  >  0. 

and  u  has  a  minimum. 

EXERCISE 

The  bottom  of  a  mural  painting  4  ft.  high  is  12  ft.  above 
the  eye  of  the  observer.  How  far  back  from  the  wall  should 
he  stand,  in  order  that  the  angle  subtended  by  the  painting  be 
as  large  as  possible  ? 

Suggestion.  Take  the  distance,  x,  of  the  observer  from  the 
wall  as  the  independent  variable,  and  express  the  angle  of 
elevation  of  the  bottom  and  the  top  of  the  painting  in  terms 
of  x. 


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REC'Q  LU     NOV    3^rfguT  8 

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